Chuyen de LTDH Do Minh Tuan Clc Cdsp Nd

8/3/2019 Chuyen de LTDH Do Minh Tuan Clc Cdsp Nd

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Th.s MINH TUN

TI LIU LUYN THI I HC

MN TON

NAM NH, NM 2009

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Li ni uTrong nhng nm gn y, thi i hc tr nn c bn hn trc rt nhiu , khngcn tnh nh cng nh bt hc sinh phi nh nhiu nhng mo rt lt vt. Mts ti liu ging dy rt hay ngy trc nh "Cc bi ging luyn thi mn Ton", "B thi tuyn sinh" ch cn li mt t gi tr thc tin ca n. Cht lc nhng ti liuny, bm st nhng thi tuyn sinh nhng nm gn y (T nm 2002-2010) cng vinhng kinh nghim trong thc tin ging dy luyn thi ca mnh (c tham kho mt s

bi ging nhng trang web dy hc) ti bin son ti liu ny mc ch chnh mnhging dy mt cch bi bn.Ti ngh rng ti liu ny s c ch i vi nhng ngi dy ton, cng nh nhng bnngp ngh cng trng i hc.Ti liu ny gm 12 chuyn (vn cn thiu)1. Phng trnh i s.2. Phng trnh lng gic.3. Phng trnh cha cn v du gi tr tuyt i.4. H phng trnh i s

5. Gii tch t hp6. Hnh phng ta 7. Gii hn8. Bt ng thc9. Hm s v th10. Hnh hc khng gian ta 11. Tch phn v ng dng12. S phcV s lng cc chuyn ln nn khng th trnh khi nhng li nh my, li tnhton sai, ... Mong cc bn lng th, mi gp xin gi v:

Th.s Minh Tun.Trng CSP Nam nh, 813 ng Trng Chinh, TP Nam nhEmail: [email protected]: 0982843882.

Chc cc bn thnh cng trong k thi i hc sp ti!

Nam nh, ngy 20 thng 06 nm 2010Tc gi

Minh Tun

Bin tp : Th.s Minh Tun Trang 2 Khoa T nhin - Trng CSP Nam nh

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1 Phng trnh i s 81.1 L thuyt v a thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.1 Phn tch a thc thnh nhn t . . . . . . . . . . . . . . . . . . 81.1.2 Tnh gi tr mt a thc, phn thc ti im l . . . . . . . . . . 9

1.2 Phng trnh bc nht . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.1 Phng php gii . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Phng trnh bc hai . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.1 Phng php gii . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 Phng trnh bc 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.1 Tnh cht ca a thc . . . . . . . . . . . . . . . . . . . . . . . . 141.4.2 a thc bc 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.3 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.5 Phng trnh bc 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5.1 Dng tng qut . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5.2 Cc dng ca phng trnh bc 4 . . . . . . . . . . . . . . . . . . 171.5.3 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.6 Du ca a thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.6.1 a thc bc 1 - bc 2 . . . . . . . . . . . . . . . . . . . . . . . . . 201.6.2 a thc - Phn thc tng qut . . . . . . . . . . . . . . . . . . . 241.6.3 Gii h bt phng trnh . . . . . . . . . . . . . . . . . . . . . . . 26

1.7 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Phng trnh lng gic 322.1 Cc kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.1.1 Cng thc lin h gia cc hm lng gic . . . . . . . . . . . . . 322.1.2 Cc cng thc ca cc gc lin h vi . . . . . . . . . . . . . . 322.1.3 Bng du ca cc hm lng gic . . . . . . . . . . . . . . . . . . 332.1.4 Bng cc gi tr lng gic . . . . . . . . . . . . . . . . . . . . . . 332.1.5 Cng thc lng gic ca tng, hiu . . . . . . . . . . . . . . . . 332.1.6 Cng thc cng lng gic . . . . . . . . . . . . . . . . . . . . . . 342.1.7 Cng thc bin i tch thnh tng . . . . . . . . . . . . . . . . . 34

2.1.8 Cng thc gc nhn i, nhn ba - Cng thc h bc . . . . . . . 34

2.1.9 Cng thc tnh sin x, cos x, tan x, cot x theo t = tanx

2. . . . . . 35

2.1.10 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35


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2.2 Cc phng trnh lng gic c bn . . . . . . . . . . . . . . . . . . . . . 362.2.1 Phng trnh sin x = m . . . . . . . . . . . . . . . . . . . . . . . 362.2.2 Phng trnh cos x = m . . . . . . . . . . . . . . . . . . . . . . . 362.2.3 Phng trnh tan x = m, cot x = m . . . . . . . . . . . . . . . . . 362.2.4 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.2.5 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 Cc phng trnh lng gic khc . . . . . . . . . . . . . . . . . . . . . . 392.3.1 Phng trnh a sin x + b cos x = c . . . . . . . . . . . . . . . . . . 392.3.2 Phng trnh ng cp cha sin v cos . . . . . . . . . . . . . . . 402.3.3 i s ha phng trnh lng gic . . . . . . . . . . . . . . . . . 412.3.4 Phng trnh i xng sin, cos . . . . . . . . . . . . . . . . . . . . 412.3.5 Phn tch thnh nhn t . . . . . . . . . . . . . . . . . . . . . . . 422.3.6 S dng bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . 432.3.7 Loi nghim khng thch hp . . . . . . . . . . . . . . . . . . . . 442.3.8 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3 Phng trnh cha cn v du gi tr tuyt i 493.1 Phng trnh cha du gi tr tuyt i . . . . . . . . . . . . . . . . . . . 49

3.1.1 Kin thc cn nh . . . . . . . . . . . . . . . . . . . . . . . . . . 493.1.2 Cc dng bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.1.3 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.2 Phng trnh cha cn thc . . . . . . . . . . . . . . . . . . . . . . . . . 513.2.1 Cc dng bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.2.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.3 Bt phng trnh cha du gi tr tuyt i . . . . . . . . . . . . . . . . 533.3.1 Dng c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.4 Bt phng trnh cha cn thc . . . . . . . . . . . . . . . . . . . . . . . 533.4.1 Dng c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.4.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.5 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4 H phng trnh i s 594.1 H phng trnh bc nht . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.1.1 H phng trnh bc nht hai n . . . . . . . . . . . . . . . . . . 594.1.2 H phng trnh bc nht ba n . . . . . . . . . . . . . . . . . . . 604.1.3 H phng trnh bc nht bn n . . . . . . . . . . . . . . . . . . 60

4.2 H phng trnh bc nht - bc hai: . . . . . . . . . . . . . . . . . . . . . 614.3 H phng trnh ng cp bc 2: . . . . . . . . . . . . . . . . . . . . . . 62

4.3.1 Phng trnh ng cp bc 2 . . . . . . . . . . . . . . . . . . . . 624.3.2 H phng trnh ng cp bc 2 . . . . . . . . . . . . . . . . . . . 63

4.4 H i xng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.4.1 H i xng loi I: . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.4.2 H i xng loi II: . . . . . . . . . . . . . . . . . . . . . . . . . . 664.5 H phng trnh tng qut . . . . . . . . . . . . . . . . . . . . . . . . . . 694.6 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71


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5 Gii tch t hp 775.1 Khi qut chung . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

5.2.1 Quy tc cng - nhn . . . . . . . . . . . . . . . . . . . . . . . . . 775.2.2 T hp - chnh hp - hon v . . . . . . . . . . . . . . . . . . . . . 78

5.2.3 Cng thc nh thc Newton . . . . . . . . . . . . . . . . . . . . . 795.3 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.4 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

6 Hnh phng ta 856.1 Vc t, im, ng thng . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.1.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.1.2 Dng bi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

6.2 ng trn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.2.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

6.2.2 Cc dng bi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936.3 Ba ng Conic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.3.1 Kin thc chung v 3 ng Conic . . . . . . . . . . . . . . . . . 1006.3.2 Elip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.3.3 Hyperbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.3.4 Parabol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

6.4 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

7 Gii hn 1267.1 Gii hn dy s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

7.1.1 Cc tnh cht c bn ca gii hn . . . . . . . . . . . . . . . . . . 1267.1.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

7.2 Gii hn hm s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1287.2.1 Gii hn c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1287.2.2 Phng php tnh gii hn . . . . . . . . . . . . . . . . . . . . . . 1297.2.3 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

7.3 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

8 Bt ng thc 135

8.1 Cc bt ng thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.2 Bt ng thc Cauchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1388.2.1 Tm min tng, max ca tch . . . . . . . . . . . . . . . . . . . . . 1388.2.2 Bt ng thc i xng . . . . . . . . . . . . . . . . . . . . . . . 1408.2.3 Cc tr c iu kin . . . . . . . . . . . . . . . . . . . . . . . . . . 144

8.3 Bt ng thc Bunhiacopxki . . . . . . . . . . . . . . . . . . . . . . . . . 1478.4 Bt ng thc Cauchy - Schwarz . . . . . . . . . . . . . . . . . . . . . . 1488.5 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

9 Hm s v th 153

9.1 Kho st v v th hm s . . . . . . . . . . . . . . . . . . . . . . . . 1539.1.1 Kin thc cn nh . . . . . . . . . . . . . . . . . . . . . . . . . . 1539.1.2 Cc bc kho st hm s . . . . . . . . . . . . . . . . . . . . . . 156


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9.1.3 Hm a thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1569.1.4 Hm phn thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

9.2 Cc tr v tim cn ca hm s . . . . . . . . . . . . . . . . . . . . . . . 1599.2.1 Quy tc tm cc i v cc tiu ca hm s . . . . . . . . . . . . 1599.2.2 Cc tr hm s . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

9.2.3 Cc bi ton tim cn . . . . . . . . . . . . . . . . . . . . . . . . 1639.2.4 Cng c kin thc . . . . . . . . . . . . . . . . . . . . . . . . . . 1669.3 Gi tr ln nht, nh nht ca hm s . . . . . . . . . . . . . . . . . . . 168

9.3.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 1689.3.2 Cc bi ton n thun . . . . . . . . . . . . . . . . . . . . . . . 1699.3.3 Bi ton gi tr ln nht, nh nht cha tham s . . . . . . . . . 1719.3.4 Phng php min gi tr ca hm s . . . . . . . . . . . . . . . . 1739.3.5 Phng php chiu bin thin . . . . . . . . . . . . . . . . . . . . 1759.3.6 Cng c kin thc . . . . . . . . . . . . . . . . . . . . . . . . . . 177

9.4 Vit phng trnh tip tuyn th . . . . . . . . . . . . . . . . . . . . . 178

9.4.1 Kin thc cn nh . . . . . . . . . . . . . . . . . . . . . . . . . . 1789.4.2 Tip tuyn vi ng cong ti im M . . . . . . . . . . . . . . . 1789.4.3 Tip tuyn vi ng cong i qua im M . . . . . . . . . . . . . 1799.4.4 Lp cc bi ton v s tip xc rt a dng . . . . . . . . . . . . 1809.4.5 Cng c kin thc . . . . . . . . . . . . . . . . . . . . . . . . . . 181

9.5 Xc nh im tha mn iu kin cho trc . . . . . . . . . . . . . . . . 1829.5.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 1829.5.2 Tm im khng thuc mi ng cong trong h y = f(x, m) . . . 184

9.6 S tng giao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

9.6.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 1869.6.2 S tng giao ca hm a thc vi trc Ox . . . . . . . . . . . . 1879.6.3 S tng giao ca hm phn thc . . . . . . . . . . . . . . . . . . 1899.6.4 Cng c kin thc . . . . . . . . . . . . . . . . . . . . . . . . . . 191

9.7 S tip xc ca 2 ng cong . . . . . . . . . . . . . . . . . . . . . . . . 1939.7.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 1939.7.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1939.7.3 Cng c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

9.8 Bin lun s nghim bng th . . . . . . . . . . . . . . . . . . . . . . . 197

9.8.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . 1979.8.2 Cc v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1989.9 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

10 Hnh khng gian ta 20810.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

10.1.1 Vct v php ton vct trong khng gian . . . . . . . . . . . . . 20810.1.2 Mt phng trong khng gian . . . . . . . . . . . . . . . . . . . . . 20910.1.3 ng thng trong khng gian . . . . . . . . . . . . . . . . . . . 21010.1.4 V tr tng i . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

10.1.5 Chm mt phng . . . . . . . . . . . . . . . . . . . . . . . . . . . 21110.1.6 Gc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21210.1.7 Khong cch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21210.1.8 Din tch, th tch . . . . . . . . . . . . . . . . . . . . . . . . . . 213


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10.1.9 Mt s dng ton v mt phng v ng thng . . . . . . . . . . 21410.1.10Mt cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

10.2 Vc t, im . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21710.3 Phng trnh mt phng . . . . . . . . . . . . . . . . . . . . . . . . . . . 22110.4 Phng trnh ng thng . . . . . . . . . . . . . . . . . . . . . . . . . . 224

11 Tch phn 22611.1 Vi phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

11.1.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22611.1.2 Cc tnh cht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22611.1.3 Bng o hm cc hm s s cp thng gp . . . . . . . . . . . 226

11.2 Nguyn hm v tch phn bt nh . . . . . . . . . . . . . . . . . . . . . 22711.2.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22711.2.2 Cc tnh cht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22711.2.3 Bng nguyn hm cc hm s s cp thng gp . . . . . . . . . 228

11.3 Cc phng php tnh tch phn . . . . . . . . . . . . . . . . . . . . . . . 22911.3.1 Php i bin s . . . . . . . . . . . . . . . . . . . . . . . . . . . 22911.3.2 Tch phn tng phn . . . . . . . . . . . . . . . . . . . . . . . . . 23111.3.3 Tch phn hm phn thc . . . . . . . . . . . . . . . . . . . . . . 232

11.4 Tch phn xc nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23411.5 ng dng ca tch phn . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

11.5.1 Tnh din tch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23511.5.2 Tnh th tch vt th trn xoay . . . . . . . . . . . . . . . . . . . 236

11.6 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

12 S phc 25912.1 Kin thc c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

12.1.1 Cc kin thc chung . . . . . . . . . . . . . . . . . . . . . . . . . 25912.1.2 Cc php ton trn s phc . . . . . . . . . . . . . . . . . . . . . 259

12.2 Cc dng bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26012.2.1 Thc hin cc php ton . . . . . . . . . . . . . . . . . . . . . . . 26012.2.2 Khai cn bc 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26012.2.3 Gii phng trnh i s v cc vn lin quan . . . . . . . . . . 26212.2.4 Biu din s phc trn mt phng . . . . . . . . . . . . . . . . . . 263

12.2.5 Chng minh ng thc t hp . . . . . . . . . . . . . . . . . . . . 26312.3 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263


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Chng 1. Phng trnh i s

Chng 1Phng trnh i s

1.1 L thuyt v a thc

1.1.1 Phn tch a thc thnh nhn t+) Nu P(x) l mt a thc bc 2 c 2 nghim x1, x2 th P(x) = a.(x x1).(x2) (a lh s bc cao nht ca P(x)).+) Tng qut: Nu P(x) l mt a thc bc n c n nghim x1, x2, , xn th

P(x) = a(x x1)(x x2) (x xn)

+) Mt a thc P(x) bt k bao gi cng phn tch thnh tch nhng a thc bc nhtv a thc bc 2 (v nghim).

V d 1.1: Phn tch cc a thc sau thnh nhn t:

a) P(x) = 2x2 5x + 2.b) P(x) = 3x2 + 12x 12c) P(x) = 4x3 4x2 7x 2.d) P(x) = 6x3 13x2 + 4x + 3

Gii: a) P(x) c a = 2, x1 = 2, x2 =1

2

nn P(x) = 2(x 2)

x 12

= (x 2)(2x 1).

b) P(x) c nghim kp x = 2 nn P(x) = 3(x 2)2.

c) P(x) c a = 4 v 2 nghim x = 12

v x = 2???

Ch : P(x) l a thc bc 3 nhng li ch c 2 nghim. Nn s c mt nghim lnghim kp. Tt nht trong trng hp ny ta dng lc Hoocne gii quyt.

Kt qu: P(x) = 4

x +1

2

2(x 2).


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1.2. Phng trnh bc nht Chng 1. Phng trnh i s

d) P(x) c a = 6 v 3 nghim x = 1, x = 13, x =

3

2

P(x) = 6(x 1).

x +1

3

x 3

2

= (x 1)(3x + 1)(2x 3).

1.1.2 Tnh gi tr mt a thc, phn thc ti im l

Cch lm: Nhp hm, s dng tnh nng CALC ca my 570ES.

V d 1.2: Tnh gi tr biu thc:

a) y = x3 3x2 x 1 ti x = 1 3 v x = 1 + 3

b) y =x2 x 1

2x + 3ti x = 3 +

2 v x = 3 2

Gii: a) x = 1 3 y = 4 + 3x = 1 +

3 y = 4 3

b) x = 3 +

2 y = 43 + 31

2

73

x = 3 2 y = 43 31

2

73

1.2 Phng trnh bc nht

1.2.1 Phng php gii

Dng ca phng trnh: ax + b = 0

Cch gii:

Vi a = 0, b = 0: Phng trnh nghim ng x R Vi a = 0, b

= 0: Phng trnh v nghim.

Vi a = 0 Phng trnh c nghim duy nht x = ba

1.2.2 Cc v d

V d 1.3: Gii v bin lun phng trnh: (m2 1)x + m 1 = 0Gii: - Nu m2 1 = 0 m = 1.+) Vi m = 1 phng trnh tr thnh: 0x + 0 = 0. Phng trnh nghim ng x R.+) Vi m =

1 phng trnh tr thnh: 0x

2 = 0. Phng trnh v nghim.

- Nu m2 1 = 0 m = 1.Phng trnh c nghim duy nht: x = 1

m + 1


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1.3. Phng trnh bc hai Chng 1. Phng trnh i s

V d 1.4: Tm im c nh ca h ng thng:(dm) : y = (m 2)x + 2m 3Gii: Gi (x0, y0) l im c nh ca (dm) y0 = (m 2)x0 + 2m 3 m m(x0 + 2) 2x0 3 y0 = 0 m

x0 + 2 = 02x0 3 y0 = 0 x0 = 20y0 = 1Vy im c nh ca h (dm) l im A(2;1)

1.3 Phng trnh bc hai

1.3.1 Phng php gii

Dng ca phng trnh: ax2 + bx + c = 0. Bin lun:

- Nu a = 0: phng trnh bc nht

- Nu a = 0: = b2 4ac hoc = b2 ac.+) Nu < 0: Phng trnh v nghim.

+) Nu = 0: Phng trnh c nghim kp x1 = x2 = b

2a= b

a+) Nu > 0: Phng trnh c 2 nghim phn bit

x1,2 = b

2a=

b a

Nhm nghim:

- Nu a + b + c = 0 th phng trnh c 2 nghim: x1 = 1, x2 =c

a

- Nu a b + c = 0 th phng trnh c 2 nghim: x1 = 1, x2 = c

a

Phn tch mt tam thc bc 2 thnh nhn t.Gi s f(x) = ax2 + bx + c c 2 nghim x1, x2 th f(x) = a(x x1)(x x2).V d: f(x) = 2x2 5x + 2 c 2 nghim x1 = 2, x2 =

1

2

nn f(x) = 2(x 2)(x 12

) = (x 2)(2x 1).

nh l Vi-et: Gi s x1, x2 l 2 nghim ca phng trnh th ta c:

x1 + x2 =

b

a

x1x2 =c

a


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nh l Vi-et o:

Nu

x + y = S

x.y = P, x, y l 2 nghim ca phng trnh:

X2 S.X+ P = 0

Du ca nghim:

- Pt c 2 nghim phn bit dng

> 0

S > 0

P > 0

- Pt c 2 nghim phn bit m

> 0

S < 0

P > 0

-

Pt c 2 nghim tri du: P < 0.- Pt c nghim dng tng ng vi phng trnh c 2 nghim dng hoc

c 2 nghim tri du

max(x1, x2) > 0

0

max(x1, x2) =

b +

2aNu a > 0

b 2a

Nu a < 0

Hoc ta c th xt 2 trng hp:

- Phng trnh c 2 nghim dng (khng cn phn bit) hoc c mt nghim

bng khng, mt nghim dng

0S > 0

P 0- Phng trnh c 2 nghim tri du P < 0.

- Phng trnh c nghim m ta lm tng t nh trn:

0S < 0

P 0

P < 0

0min(x1, x2) < 0

min(x1, x2) =

b

2aNu a > 0

b + 2a

Nu a < 0

So snh nghim vi mt s:

- (x1, x2) a.f() < 0.

- / [x1, x2] 0a.f() > 0

- x1 < x2 <

> 0

a.f() > 0

S/2 <


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- x1 > x2 >

> 0

a.f() > 0

S/2 >

V d 1.5: Gii cc phng trnh sau:

a) x2 5x + 4 = 0b) x2 2x 3 = 0

Gii: a) a + b + c = 0 phng trnh c nghim x1 = 1, x =c

a= 4

b) a b + c = 0 phng trnh c 2 nghim x1 = 1, x = c

a= 3

V d 1.6: Gii v bin lun phng trnh sau: (m 1) x2

(2m + 1) x + m 5 = 0.Gii: +) TH 1: Nu m 1 = 0 m = 1 thay vo phng trnh ta c:3x 4 = 0 x = 3

4.

+) TH 2: Nu m 1 = 0 m = 1. = (2m + 1)2 4 (m 1) (m 5) = 28m 19- Nu > 0 m > 19

28c phng trnh c 2 nghim phn bit:

x1,2 =

2m + 1

28m

19

2 (m 1)

- Nu = 0 m = 1928

c nghim kp:

x1 = x2 =2m + 1

2 (m 1) =2.

19

28+ 1

2

19

28 1 = 11

3

- Nu < 0 m < 1928

: Phng trnh v nghim.

V d 1.7: Cho phng trnh x2 (m 1) x + 2m 5 = 0a) Tm m phng trnh c 2 nghim phn bit x1, x2.

b) Tm h thc lin h gia x1, x2 khng ph thuc vo m.

c) Lp phng trnh bc 2 nhn 2x1 + x2 l nghim.

Gii: a) = (m 1)2 4 (2m 5) = m2 2m + 1 8m + 20 = m2 10m + 21.Phng trnh c 2 nghim phn bit > 0 m2 10m + 21 > 0

m > 7

m < 3


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b) S = x1 + x2 = m 1, P = x1.x2 = 2m 5. Do 2S P = 2(m 1) (2m 5) = 3H thc lin h x1, x2 khng ph thuc m l : 2(x1 + x2) x1.x2 = 3

c) t u = 2x1 + x2, v = x1 + 2x2. Do :

u + v = 3(x1 + x2) = 3.(m 1) = 3m 3,u.v = (2x1 + x2)(x1 + 2x2) = 2x21 + 5x1.x2 + x22 = 2(x1 + x2)2 + x1.x2

= 2(m 1)2 + 2m 5 = 2m2 2m 3Do u, v l 2 nghim ca phng trnh:

X2 (3m 3)X+ 2m2 2m 3 = 0

V d 1.8: Cho phng trnh: x2 (m + 1)x + m + 94

= 0

1. Tm m phng trnh c 2 nghim x1, x2.

2. Tm m phng trnh c 2 nghim dng.

3. Tm m phng trnh c nghim dng.

4. Tm m phng trnh c 2 nghim tha mn x1 < 1 < x2.

5. Tm m phng trnh c 2 nghim tha x1 x2 < 2

Gii: a) = (m + 1)2 4(m + 94

) = m2 + 2m + 1 4m 9 = m2 2m 8

Phng trnh c 2 nghim phn bit > 0 m2 2m 8 > 0 m > 4m < 2

b) Phng trnh c 2 nghim dng

> 0

S = m + 1 > 0

P = m + 9/4 > 0

m > 4

m < 2m > 1m > 9

4 m > 4

c) Phng trnh c nghim dng 0max(x1, x2) > 0

m

2 2m 8 0m + 1 +

m2 2m 82

> 0 m2 2m 8 > m 1

m 1 < 0m2 2m 8 0

m 1 0

m

2

2

m 8

>(m

1)2

m > 1m 4m 2

m 1m < 9/4

m 4m < 9/4

d) Phng trnh c nghim x1 < 1 < x2 a.f(1) < 0 1. (1 (m + 1) + m + 9/4 < 0) 9/4 < 0 (V nghim)


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1.4. Phng trnh bc 3 Chng 1. Phng trnh i s

e) Phng trnh c nghim tha mn x1 x2 < 2

0a.f(2) > 0

S/2 < 2

m 4 m 2m < 17/4

m < 3

4 m < 17/4m

2

1.4 Phng trnh bc 3

1.4.1 Tnh cht ca a thc

nh l Berzout: Cho P(x) l mt a thc bt k. Khi vi mi x0, a thcP(x) chia a thc x x0 c s d l P(x0).

H qu: Nu x0 tha mn P(x0) = 0 th P(x)... x x0. Lc Hoocne: Gi s P(x) = anxn + an1xn1 + + a1x + a0.

an an1 a1 a0x0 bn bn1 b1 b0

bn = an, bn1 = bn.x0 + an1, bn2 = bn1.x0 + an2, , b0 = b1.x0 + a0.P(x) = (x x0)(bnxn1 + bn1xn2 + + b2x + b1) + b0.

Nu P(x)... x x0 th b0 = 0 v P(x) = (x x0)(bnxn1 + bn1xn2 + + b2x + b1).

1.4.2 a thc bc 3

Dng ax3 + bx2 + cx + d = 0 (1).

Cch gii :

- Nhm nghim : S dng my tnh nhm mt nghim x0 no .

- Dng lc Hooc-ne phn tch a thc trn thnh nhn t :

P(x) = (x x0).Q(x). Q(x) l mt a thc bc 2. nh l Viet: Gi s x1, x2, x3 l 3 nghim ca phng trnh (1).

x1 + x2 + x3 = b

a

x1x2 + x2x3 + x3x1 =c

a

x1x2x3 =

d

a


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nh l Viet o: Gi s x,y,z l 3 s tha mn

x + y + z = m

xy + yz + zx = n

xyz = p

Khi x,y,z l 3 nghim ca phng trnh : X3 mX2 + nXp = 0

1.4.3 Cc v d


a) 2x3 x2 + x + 4 = 0.b) x3 4x + 3 = 0.Gii: a) Dng my tnh ta thy c mt nghim l : x =

1.

Dng lc Hooc-ne ta c:

2 1 1 41 2 3 4 0

Phng trnh (x + 1) (2x2 3x + 4) = 0

x = 12x2 3x + 4 = 0Phng trnh v nghim x = 1.

b) Dng my tnh ta nhm c nghim x = 1.1 0 4 3

1 1 1 3 0

Phng trnh (x 1) (x2 + x 3) = 0

x = 1

x2 + x 3 = 0 x = 1

x =1 13

2

V d 1.10: Cho phng trnh 2x3 3x2 5x + 5 = 0a) Chng minh phng trnh c 3 nghim x1, x2, x3 phn bit.

b) Tnh P = 3 (x21 + x22 + x

23) 2 (x31 + x32 + x33).

Gii: a) t f(x) = 2x3 3x2 5x + 5. Ta c f(2) = 13, f(1) = 5, f(1) = 1,f(3) = 17.

Ta c f(2).f(1) < 0 nn tn ti x1 (2; 1) sao cho f(x1) = 0.f(

1).f(1) < 0 nn tn ti x2

(

1;1) sao cho f(x2) = 0.

f(1).f(3) < 0 nn tn ti x3 (1;3) sao cho f(x3) = 0.Do ta c f(x1) = f(x2) = f(x3) = 0 v x1 < x2 < x3 nn phng trnhf(x) = 0 c 3 nghim phn bit.


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b) Theo nh l Viet ta c:

x1 + x2 + x3 =3

2

x1x2 + x2x3 + x3x1 = 5

2

x1x2x3 = 52

x21 + x22 + x

23 = (x1 + x2 + x3)

2 2 (x1x2 + x2x3 + x3x1) =29

4

x31 + x32 + x

33 = (x

31 + x

32 + x

33 3x1x2x3) + 3x1x2x3

= (x1 + x2 + x3) (x21 + x

22 + x

23 (x1x2 + x2x3 + x3x1)) + 3x1x2x3

=3

2

29

4+

5

2

+ 3.

5

2

=

57

8

Do ta c P = 3.29

4 2.57

8 =

15

2 .

V d 1.11: Gii h phng trnh:x + y + z = 2

x2 + y2 + z2 = 6

x3 + y3 + z3 = 8

Gii: Phng trnh tng ng vi

x + y + z = 2

(x + y + z)2 2 (xy + yz + zx) = 6(x

3

+ y3

+ z3

3xyz) + 3xyz = 8

x + y + z = 2

22 2 (xy + yz + zx) = 6(x + y + z) (x2 + y2 + z2 xy yz zx) + 3xyz = 8

x + y + z = 2

xy + yz + zx = 12 (6 + 1) + 3xyz = 8

x + y + z = 2

xy + yz + zx = 1xyz = 2

T ta c x,y,z l 3 nghim ca phng trnh:

X3

2X2

X+ 2 = 0 X = 1X = 1X = 2

Vy h c 6 nghim phn bit

(1;1;2) , (1;2;1) , (1; 1;2) , (1;2; 1) , (2; 1;1) , (2;1; 1)

1.5 Phng trnh bc 4

1.5.1 Dng tng qut

ax4 + bx3 + cx2 + dx + e = 0 (a = 0)


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Hng gii:

- Dng tnh nng SOLVE hoc TABLE ca my tnh fx-570ES, fx-500ES nhm nghim ca phng trnh, sau dng lc Hooc-ne phn tchthnh phng trnh bc 3 v gii tip nh trn.

-

Tuy nhin mt s trng hp cch gii trn tr nn v hiu hoc qu phctp khng cn thit, nhng trng hp c cch gii ring bit.

1.5.2 Cc dng ca phng trnh bc 4

Phng trnh trng phng : ax4 + bx2 + c = 0.

Cch gii: t t = x2 0. Phng trnh tr thnh : at2 + bt + c = 0. Phn tch thnh nhn t:

Cch gii: Bit c mt nghim, hoc dng cch nhm, s dng hng ng thc

phn tch thnh nhn t, quy v phng trnh bc thp hn.

Phng trnh i xng: ax4 + bx3 + cx2 + dx + e = 0 tha mnd

b

2=

e

a

Cch gii: Xt x = 0 thay vo phng trnh xem c tha mn khng?

Vi x = 0. Chia c 2 v ca phng trnh cho x2 ta c:

ax2 + bx + c + dx

+ ex2

= 0 ax2 + eax2+ bx + d

bx+ c = 0

t t = x +b

dx()

t2 = x2 + b2

d2x2+

2d

b= x2 +

e

ax2+

2d

b.

Phng trnh tr thnh: a

t2 2d

b

+ bt + c = 0

Gii phng trnh bc 2 n t. Sau thay vo () tm x. Phng trnh dng (x + a) (x + b) (x + c) (x + d) = e sao cho a + b = c + d.

Cch gii: Phng trnh (x2 + (a + b) x + ab) (x2 + (c + d) x + cd) = et t = x2 + (a + b) x = x2 + (c + d) x ()Thay vo phng trnh ta c:

(t + ab) (t + cd) = e

Gii phng trnh bc 2 n t sau thay vo () tm x.


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Phng trnh dng (x + a) (x + b) (x + c) (x + d) = ex2 sao cho ab = cd.

Cch gii: ging cch gii phng trnh i xng.

Nu x = 0: ta c abcd = 0.

Nu x = 0: Phng trnh (x2 + (a + b) x + ab) (x2 + (c + d) x + cd) = ex2.

x + abx

+ a + b .x + cdx

+ c + d = et t = x +

ab

x= x +

cd

x(). Phng trnh tr thnh:

(t + a + b) (t + c + d) = e

Gii phng trnh bc 2 ta tm c t. Thay vo () tm x.

1.5.3 Cc v d

V d 1.12: Gii phng trnh 2x4 x2 3 = 0Gii: t t = x2 0. Phng trnh tr thnh :

2t2 t 3 = 0 t = 1 (loi)

t =3

2

t = 32

x2 = 32

x =

6

2


a) 8x4 + 16x3 8x2 91x 42 = 0.

b) x4 4x3 + 4x2 16 = 0.c) x4 4x 1 = 0.Gii: a) Dng my tnh ta nhm c mt nghim l x = 2.

Dng lc Hooc - ne ta c:

8 16 8 91 422 8 32 56 21 0

Phng trnh

(x

2)(8x3 + 32x2 + 56x + 21) = 0.

Tip tc ta nhm c 1 nghim l x = 12

. Theo lc Hooc - ne ta c:

8 32 56 21

12

8 28 42 0

Phng trnh (x 2)

x +1

2

(8x2 + 28x + 42) = 0

x = 2

x = 12

8x2 + 28x + 42 = 0V nghim


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b) Phng trnh (x2 2x)2 42 = 0 (x2 2x 4) (x2 2x + 4) = 0

x2 2x 4 = 0x2 2x + 4 = 0 V nghim x = 1 5

c) Phng trnh x4 + 2x2 + 1 2 (x2 + 2x + 1) = 0

(x2

+ 1)2

2 (x + 1)2 = 0 x2 + 1 2 (x + 1) x2 + 1 + 2 (x + 1) = 0 x2 2x + 1 2 x2 + 2x + 1 + 2 = 0

x2 2x + 1 2 = 0x2 +

2x + 1 +

2 = 0

V nghim

x =

2

2 + 422

.

V d 1.14: Gii cc phng trnh sau:a) x4 + 4x3 x2 + 8x + 4 = 0.b) 2x4 3x3 3x2 + 3x + 2 = 0.Gii: a) Vi x = 0, phng trnh tr thnh 2 = 0 (v l). Vy x = 0.

Chia c 2 v phng trnh cho x2 ta c

x2 + 4x 1 + 8x

+4

x2= 0

x2 +

4

x2

+ 4

x +

2

x

1 = 0

t t = x +2

x t2 = x2 + 4

x2+ 4

Phng trnh tr thnh : t2 4 + 4t 1 = 0

t2 + 4t 5 = 0

t = 1

t = 5

+) Vi t = 1: x +2

x= 1 x2 x + 2 = 0 v nghim

+) Vi t = 5: x +2

x = 5 x2

+ 5x + 2 = 0 x =5 17

2

b) x = 0 khng l nghim ca phng trnh nn chia c 2 v ca phng trnh chox2 = 0 ta c

2x2 3x 3 + 3x

+2

x2= 0 2

x2 +

1

x2

3

x 1x

3 = 0

t t = x 1x

t2 = x2 + 1x2

2, thay vo phng trnh ta c:

2 (t2 + 2) 3t 3 = 0 2t2 3t + 1 = 0 t = 1t = 12

+) Vi t = 1: x 1x

= 1 x2 x 1 = 0 x = 1

5

2


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1.6. Du ca a thc Chng 1. Phng trnh i s

+) Vi t =1

2: x 1

x= 1

2 2x2 x 2 = 0 x = 1

17

4.

V d 1.15: Gii phng trnh sau : x (x + 1) (x 3) (x 2) = 2Gii: Phng trnh

(x2

2x) (x2

2x

3) =

2

t t = x2 2x. Phng trnh tr thnh:t (t 3) = 2 t2 3t + 2 = 0

t = 1

t = 2

+) Vi t = 1: x2 2x = 1 x2 2x 1 = 0 x = 1 2.+) Vi t = 2: x2 2x = 2 x2 2x 2 = 0 x = 1 3.V d 1.16: Gii phng trnh (x 2) (x + 3) (x 1) (x + 6) = 21x2

Gii: Phng trnh (x2 + x 6) (x2 + 5x 6) = 21x2 Do x = 0 khng l nghimca phng trnh nn chia c 2 v ca phng trnh cho x2

= 0 ta c:

x + 1 6x

x + 5 6

x

= 21

t t = x 6x

thay vo phng trnh ta c:

(t + 1) (t + 5) = 21 t2 + 6t + 5 = 21 t2 + 6t 16 = 0

t = 8t = 2

+) Vi t = 8: x 6

x = 8 x2 + 8x 6 = 0 x = 4 22

+) Vi t = 2: x 6x

= 2 x2 2x 6 = 0 x = 1 7

1.6 Du ca a thc

1.6.1 a thc bc 1 - bc 2

Dng: P(x) = ax + b (a = 0). Ta c bng xt du:

x ba

+P(x) sign(a) 0 +sign(a)

sign(a) l du ca a.

Dng P(x) = ax2 + bx + c (a = 0). = b2 4ac. Ta c cc trng hp sau:+) < 0: Du ca a thc l:

x +P(x) +sign(a)


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+) = 0: Du ca a thc l:

x b2a

+P(x) +sign(a) 0 +sign(a)

+) > 0: P(x) c 2 nghim phn bit x1, x2. Du ca a thc l :

x x1 x2 +P(x) +sign(a) 0 sign(a) 0 +sign(a)

Ch : Nu P(x) l mt a thc bc 2 ta lun c:

P(x) > 0 x R

< 0

a > 0

P(x) < 0 x R < 0a < 0 P(x) 0 x R

0a > 0

V d 1.17: Xt du ca cc biu thc sau:

a) P(x) = 2x + 3b) P(x) = x2 + 4x 5

c) P(x) = 4x2 12x + 9d) P(x) = x2 x 6e) P(x) = 2x2 + 3x + 2

Gii: a) P(x) = 0 x = 32, a = 2 < 0. Do du ca P(x) l:

x 32

+P(x) +

0

b) = 4 < 0, a = 1 < 0, ta c du ca P(x) l:

x +P(x) +

c) = 0, a = 4 > 0 v du ca P(x) l:

x 3

2 +P(x) + 0 +

d) > 0, x1 = 3, x2 = 2, a = 1 > 0. Do du ca P(x) l:


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x 2 3 +P(x) + 0 0 +

e) > 0, x1 = 1

2, x2 = 2, a = 2 < 0. Do du ca P(x) l:

x 12 2 +

P(x) 0 + 0

Ch : Trong mt bi ton thng thng khng ai li hi trc tip du ca mta thc m thng hi cc cu hi v gii bt phng trnh. Chng ta cn xt duca cc a thc tng ng t tm thy c tp nghim ca bt phng trnh.Chng hn:

2x + 3 > 0 th tp nghim S =

; 32

.

2x + 3 0 th tp nghim S =

3

2; +

x2 + 4x 5 > 0 th S = x2 + 4x 5 0 th S = x2 + 4x 5 < 0 th S = R x2 + 4x 5 0 th S = R

4x2 12x + 9 > 0 th S = R\

3

2

.

4x2 12x + 9 0 th S = R 4x2 12x + 9 < 0 th S =

4x2 12x + 9 0 th S =3

2

x2 x 6 > 0 th S = (; 2) (3;+) x2 x 6 0 th S = (; 2] [3;+) x2 x 6 < 0 th S = (2;3) x2 x 6 0 th S = [2;3]

2x2 + 3x + 2 > 0 th S =

12

; 2


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2x2 + 3x + 2 0 th S =1

2; 2

2x2 + 3x + 2 < 0 th S =

; 12

(2;+)

2x2 + 3x + 2 0 th S =

; 12

[2;+)

V d 1.18: Cho tam thc bc 2: P(x) = (3m 1)x2 2(m + 1)x + 2a) Tm m P(x) = 0 c 2 nghim phn bit x1, x2.

b) Tm m f(x) =

P(x) xc nh trn R.

c) Tm m f(x) = ln P(x) xc nh trn R.

Gii: a) Ta c = (m + 1)2 2(3m 1) = m2 4m + 3. phng trnh P(x) = 0 c 2 nghim phn bit

> 0a = 0

m2 4m + 3 > 03m 1 = 0

m > 3

m < 1

m = 13

m > 3 m < 1m = 1

3

b) f(x) =

P(x) xc nh trn R P (x) 0 x R.

+) Nu 3m 1 = 0 m = 13

khi :

P(x) = 83

x + 2, r rng P(3) = 6 < 0 nn P(x) 0 khng ng vi mi x R.+) Nu 3m 1 = 0. Khi P(x) l mt a thc bc 2 do :

P(x) 0 x R

0a > 0

m2 4m + 3 03m 1 > 0

1 m 3

m >1

3 1

m

3

Kt lun: 1 m 3 tha mn iu kin bi ton.c) f(x) = ln P(x) xc nh trn R P (x) > 0 x R

+) Nu 3m 1 = 0 m = 13

khi :

P(x) = 83

x + 2, r rng P(3) = 6 < 0 nn P(x) > 0 khng ng vi mi x R.+)Nu 3m

1

= 0. Khi P(x) l mt a thc bc 2 do :

P(x) > 0 x R > 0

a > 0


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m2 4m + 3 > 03m 1 > 0

1 < m < 3m > 13

1 < m < 3

Kt lun: 1 < m < 3 tha mn iu kin bi ton.

1.6.2 a thc - Phn thc tng qut

a thc bc n: P(x) = anxn + an1xn1 + + a1x + a0.

Phn thc hu t: f(x) =P(x)

Q(x). Trong P(x), Q(x) l cc a thc.

nh l 1.1 (nh l c bn i s). Cho P(x) l mt a thc bt k th P(x) s phntch thnh tch cc a thc bc nht v a thc bc 2. Hn th na cc a thc bc u

c < 0

nh l 1.2. Cho f(x) =P(x)

Q(x)l mt phn thc hu t no . Khi trn khong

gia 2 khng im lin tip ca f(x), hmf(x) ch mang mt du.Khng im l nhng gi tr ca x m P(x) = 0, hoc Q(x) = 0.

nh l 1.3. Cho f(x) =P(x)

Q(x), khi binx chy qua khng im bi chn thf(x) khng

i du, cn qua khng im bi l th f(x) i dux0 l khng im bi chn (t. l) caf(x) nu n l mt khng im caf(x) vf(x)

cha nhn t (x x0)k vi k Z vk l s chn (t. l). Cch xt du phn thc hu t: xt du ca mt phn thc hu t ta

phn tch cc a thc ca t v mu thnh tch cc a thc bc 1, v bc 2. Cca thc bc 2 nu c 0 ta phn tch chng thnh tch cc a thc bc 1, cnnu < 0 ta thay th a thc bi h s ca hng t bc 2. Cui cng ta cphn thc ch cn tch cc a thc bc 1. Dng cc nh l trn xt du.

V d 1.19: Xt du ca biu thc sau:

a) f(x) = (x + 1)2.(x 2)3. (2x 1)

(x2 + 2x + 2)7(2x + 1)5 (1 4x) (x2 + 4x 5)3

b) f(x) =

x2 4x + 3. (2x2 5x + 2)Gii: a) Gii cc phng trnh:

+) x + 1 = 0 x = 1.+) x 2 = 0 x = 2.

+) 2x

1 = 0

x =1

2+) x2 + 2x + 2 = 0, < 0, a = 1.

+) 2x + 1 = 0 x = 12


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+) 1 4x = 0 x = 14

+) x2 + 4x 5 = 0, < 0, a = 1.Do du ca f(x) l du ca g(x) vi:

g(x) = (x + 1)

2

(x 2)3

(2x 1)17(2x + 1)5 (1 4x) (1)3 = (x + 1)

2

(x 2)3

(2x 1)(2x + 1)5 (1 4x)

Cc khng im x = 1;2; 12

; 12

;1

4. Ta c bng du:

x 1 12

1

4

1

2 2 +g(x) + 0 + + 0 0 +f(x) + 0 + + 0 0 +

b) TX: x2 4x + 3 0 x 3x 1

Ta c bng xt du:

x 12

1 2 3 +x2 4x + 3 + + 0 0 +

2x2 5x + 2 + 0 0 + +f(x) + 0 0 0 +

Dng kt qu ca v d trn ta c th gii c cc bt phng trnh:

+)(x + 1)2.(x 2)3. (2x 1)

(x2 + 2x + 2)7(2x + 1)5 (1 4x) (x2 + 4x 5)3 0

c tp nghim l: S =

; 1

2

1

4;

1

2

[2;+)

+)(x + 1)2.(x 2)3. (2x 1)

(x2 + 2x + 2)7(2x + 1)5 (1 4x) (x2 + 4x 5)3 > 0

c tp nghim l: S = (; 1) 1; 12 1

4; 1

2 (2;+)

+)(x + 1)2.(x 2)3. (2x 1)

(x2 + 2x + 2)7(2x + 1)5 (1 4x) (x2 + 4x 5)3 0

c tp nghim l: S =

1

2;

1

4

1

2; 2

{1}

+)(x + 1)2.(x 2)3. (2x 1)

(x2 + 2x + 2)7(2x + 1)5 (1 4x) (x2 + 4x 5)3 < 0

c tp nghim l: S = 12; 14 12; 2+)

x2 4x + 3. (2x2 5x + 2) 0 c tp nghim S =

; 1

2

[3;+) {1}


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+)

x2 4x + 3. (2x2 5x + 2) > 0 c tp nghim S =

; 12

(3;+)

+)

x2 4x + 3. (2x2 5x + 2) 0 c tp nghim S =

1

2; 1

{3}

+) x2

4x + 3. (2x2

5x + 2) < 0 c tp nghim S = 12; 11.6.3 Gii h bt phng trnh

Kin thc cn nh

gii c h cc bt phng trnh ta phi bit c cc thao tc ly giao v ly hpca 2 tp hp. C th nh sau:

Ly giao: phi ng thi thuc 2 (hay nhiu) tp hp. Biu din tng tp hp mt

trn trc s, xa nhng phn khng thuc tp hp i. Phn cn trng (cha bgch) chnh l tp hp cn tm.

Ly hp: ch cn thuc mt trong 2 (hay nhiu) tp hp. Biu din tng tp trntrc s: xa nhng phn khng thuc tp . Hp ca n tp hp l nhng tpkhng b xa khng qu n 1 ln.

Hc sinh thng gp kh khn khi ly hp 2 tp hp, thng ch lm tt vi trnghp ly giao.

V d 1.20: Tnh tp hp X trong cc trng hp sau:

a) X = A B C vi A = (2;1] [2;+), B = [3;0), C = (; 1].b) X = A B vi A = (3;3) v B = [1; 5].Gii: a) X = (2; 1]b) X = (3;5]

V d 1.21: Gii h phng trnh sau:

a) x

2 3x + 2 > 02x2 3x 2

3x + 1 0

b)

x2 + 3x 4 > 02x + 12x2 5x + 2 0

Gii: a) x

; 12

13

; 1

b) x

; 12

1

2; +


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1.7. Bi tp Chng 1. Phng trnh i s

1.7 Bi tp

Bi 1.1: Gii v bin lun phng trnh sau theo m:

a) (m2 + m)x + m = 2x + m2.

b) mx + m 2xx 2 = 0

Bi 1.2: Cho phng trnh: (m2 4)x2 + 2(m + 2)x + 1 = 0a) Tm m phng trnh c nghim.

b) Tm m phng trnh c nghim duy nht.

Bi 1.3: Bin lun s nghim ca phng trnh sau theo m:

x

3

m (x + 2) + x2

+ 4 = 0

Bi 1.4: Gii v bin lun phng trnh sau theo m:

a) x2 mx + 3m 8 = 0.b) x2 mx + m2 3 = 0.c) (m 2)x2 2(m + 1)x + m = 0.

Bi 1.5: Cho phng trnh 2x2 + 7x + 1 = 0. Gi x1, x2 l 2 nghim ca phng trnh.

a) Tnh gi tr ca cc biu thc:

A = 2

x21 + x22

3 x31 + x32B = x21x

32 + x

31x

22 + 2 |x1 x2|

b) Tm phng trnh vi h s nguyn nhn x1 + 2x2, 2x1 + x2 l nghim.

c) Tm phng trnh vi h s nguyn nhn x1 +1

x2l nghim.

Bi 1.6: Tm m phng trnh 3x2 + 4(m 1)x + m2 4m + 1 = 0 c 2 nghim phnbit x1, x2 tha mn

1

x1+

1

x2=

1

2.(x1 + x2)

Bi 1.7: Cho phng trnh x2 mx + (m 2)2 = 0.a) Tm m phng trnh c 2 nghim phn bit.

b) Tm gi tr ln nht, nh nht ca biu thc: F = x1x2 + 2x1 + 2x2.

Bi 1.8: Tm m phng trnh : x2 (2m + 1)x + m2 + 1 = 0 c 2 nghim x1, x2 thamn x1 = 2x2.

Bi 1.9: Tm m phng trnh x2 mx + 3m 8 = 0.


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a) C 2 nghim phn bit.

b) C 2 nghim dng.

c) C 2 nghim m.

d) C 2 nghim tri du.

e) C nghim dng.

Bi 1.10: Bin lun theo m s nghim ca phng trnh :

a) x4 2mx2 + m + 12 = 0.b) mx4 (2m 5)x2 + m + 1 = 0.c) 4x (m + 2)2x+1 + m2 4 = 0.

Bi 1.11: Gii cc phng trnh sau:a) 2x3 8x2 + x + 14 = 0.b) x3 x2 + 2 = 0.c) 3x3 4x2 5x + 6 = 0.

Hng dn. a) x = 2.

b) x = 1.

c) x = 1, x = 1 736

.

Bi 1.12: Gii cc phng trnh sau:

a) 2x4 5x2 7 = 0.b) x4 5x2 + 6 = 0.c) x4 5x3 12x2 + 15x + 9 = 0d) 2x4 + 3x3

16x2

17x + 12 = 0

e) x4 5x3 + 8x2 10x + 4 = 0.f) x4 2x3 5x2 + 2x + 1 = 0.g) (x2 2x)(x2 + 4x + 3) = 7.h) (x2 3x + 2)(x2 + 9x + 18) = 12x2.i) x4 4x3 + 8x 12 = 0.

j) x4

+ 2x3

2x2

9x 6 = 0.k) x4 3x3 2x2 + 5x + 3 = 0.l) x4 5x3 12x2 + 15x + 9 = 0.


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m) 2x4 + 3x3 16x2 17x + 12 = 0.

Hng dn. a) x =

14

2

b) x = 2 2.

c) x = 1 52

, x = 3 132

.

d) x = 1 29

2.

e) x = 3, x = 2, x = 7

73

2

f) x = 1 7.

g) x = 2, x = 1.Bi 1.13: Cho bt phng trnh: (m + 1)x + m + 2 > 0.

a) Gii v bin lun bt phng trnh.

b) Tm m bt phng trnh nghim ng vi mi x 2.Bi 1.14: Tm a h bt phng trnh sau v nghim:

x2 + 7x 8 < 0a2x + 1 > 3 + (3a

2) x

Bi 1.15: Tm m cc bt phng trnh sau nghim ng vi mi x:

a) (m 1)x2 (2m + 1)x + m + 3 > 0.b) x2 2mx + m + 12 > 0.c) x2 2x + m|x 1| + m2 2 > 0.

d)2

3 x

2 mx + 1x2

x + 1

32.

Bi 1.16: Gii cc bt phng trnh sau:

a) 4x2 + 12x 9 < 0.b) 2x2 5x + 4 > 0.c) x2 4x + 5 0.

Hng dn. a) x = 32.

b) x R.c) x [5;1].Bi 1.17: Tm a biu thc

(a + 1) x2 2 (a 1) x + 3a 3 c ngha vi mi x.


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Bi 1.18: Gii v bin lun bt phng trnh sau theo tham s m:

a) x2 mx + 2m 3 0.b) (3m 8)x2 + mx + 1 < 0.


a) x3 3x2 + 2 > 0.b) x4 + 4x3 + 4x2 25 > 0.c) x4 4x 1 < 0.d) x(x + 1)(x + 2)(x + 3) > 24.

e) x2 + x 3x2 + x

2

0

f)1

x + 1+

1

x + 2+

1

x + 3 0

g) x4 4x3 + 8x 5 < 0.h) x4 2x3 + 6x 9 < 0.

i)3x + 7

x2 x 2 5.

Hng dn. a) x 1

3; 1 1 +

3; +

b) x (; 1 6) (1 + 6; +)

c) x

2

2 + 422

;

2 +

2 + 422

d) x (; 4) (1;+).

e) x

1 132

; 2

1; 1 + 13

2

f) x

3; 2 13

2; 2 + 13

(1; +)

g) x 1 6; 1 1;1 + 6h) x (3; 3)

i) x (; 1) 3

5; 1

(2;+)


a)

x2 2x 3.x2.(2x + 1)3

(3x 8) .(x 2)5 0


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b)(x 1)4. (x + 2)(x + 3) . (2x 5) < 0

Hng dn. a) x [3;+) {1}

b) x

(

;

3)

(

2;1)1;

5

2Bi 1.21: Gii h bt phng trnh sau:

a)

(2x2 + x 2) (x 2) > 0

x2 2x 8x + 2

0

b)

x3 x2 x 2 0(x2 + 2x 3) .x + 1 0

Hng dn. a) x 1 174

; 1 + 174

(2;4]b) x [1;+) {1}Bi 1.22: Tnh gi tr biu thc trong cc trng hp sau:

a) y = x3 3x2 2x 1 khi x = 1 2.

b) y =x2 x 4

2x + 1khi x = 2 +

3.


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Chng 2. Phng trnh lng gic

Chng 2Phng trnh lng gic

2.1 Cc kin thc c bn

2.1.1 Cng thc lin h gia cc hm lng gicsin2 + cos2 = 1

1

cos2 = 1 + tan2

1

sin2 = 1 + cot2

tan cot = 1

Nhn xt:

Nu bit mt trong cc gi tr lng gic th ta c th tnh c cc gi tr lnggic cn li.

sin , cos [1;1]

2.1.2 Cc cng thc ca cc gc lin h vi

v :

cos(

) =

cos sin(

) = sin

tan( ) = tan cot( ) = cot

v :

cos() = cos sin() = sin tan() = tan cot() = cot

v + :

cos( + ) =

cos sin( + ) =

sin

tan( + ) = tan cot( + ) = cot

v

2 :


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2.1. Cc kin thc c bn Chng 2. Phng trnh lng gic

cos(

2 ) = sin sin(

2 ) = cos

tan(

2 ) = cot cot(

2 ) = tan

2.1.3 Bng du ca cc hm lng gic

I II III IVcos + +sin + + tan + + cot + +

2.1.4 Bng cc gi tr lng gic

00 300 450 600 900 1200 1350 1500 1800

0

6

4

3

2

2

3

3

4

5

6

cos 1

3

2

2

2

1

20 1

2

2

2

3

21

sin 01

2

2

2

3

2

1

3

2

2

2

1

2

0

tan 01

31

3 3 1 1

30

cot 3 1 13

0 13

1 3

2.1.5 Cng thc lng gic ca tng, hiu

cos(a b) = cos a cos b sin a sin bsin(a b) = sin a cos b cos a sin btan(a b) = tan a tan b

1 tan a tan bcot(a b) = cot a cot b 1

cot b cot a


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2.1.6 Cng thc cng lng gic

cos a + cos b = 2 cosa + b

2cos

a b2

cos a cos b = 2sin a + b2

sina b

2

sin a + sin b = 2 sin a + b2

cos a b2

sin a sin b = 2 cos a + b2

sina b

2

tan a tan b = sin(a b)cos a cos b

cot a cot b = sin(b a)sin a sin b

H qu:

tan a + cot a =2

sin2atan a cot a = 2cot2asin a + cos a =

2 sin(x +

4) =

2cos(x

4)

sin a cos a = 2sin(x 4

) = 2 cos(x + 4

)

2.1.7 Cng thc bin i tch thnh tng

cos a cos b =1

2[cos(a + b) + cos(a b)]

sin a sin b = 12

[cos(a + b) cos(a b)]

sin a cos b =1

2[sin(a + b) + sin(a b)]

2.1.8 Cng thc gc nhn i, nhn ba - Cng thc h bc

cos2x = cos2 x sin2 x = 2 cos2 x 1 = 1 2sin2 xsin2x = 2 sin x cos x

tan2x =2tan x

1

tan2 x

cot2x = cot2 x 1

2cot x

cos2 x =1 + cos 2x

2


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sin2 x =1 cos2x

2

cos3x = 4 cos3 x 3cos xsin3x = 3 sin x 4sin3 x

tan3x =3tan x

tan3 x

1 3tan2 xcot3x =

3cot x cot3 x1 3cot2 x

2.1.9 Cng thc tnh sin x, cos x, tan x, cot x theo t = tanx

2

sin x =

2t

1 + t2, cos x =

1

t2

1 + t2, tan x =

2t

1 t2, cot x =1

t2

2t

2.1.10 Bi tp

Bi 2.1: Cho l gc sao cho sin = 13, vi thuc gc phn t th III.

a) Tnh tan , cos , cot .

b) Tnh cos( 52

), tan(72

).Bi 2.2: Tnh gi tr biu thc sau:

a) A =1

sin2 200+

1

cos2 400 tan2 700 cot2 500.

b) B = sin2 10 + sin2 30 + + sin2 890.c) C = cos 20 + cos 40 + + cos 1800.

Bi 2.3: Tnh gi tr biu thc:

a) A = cos 8

+ cos 58

+ sin 98

+ sin 58

b) B = cos8 + tan

8

Rt gn biu thc:

a) A = cos x cos2x cos4x cos8x bit x = k (k Z).b) B = sin x sin(

3 x)sin(

3+ x).

Bi 2.4: Tnh gi tr biu thc:

a) A = cos6 x + sin6 x.

b) B = cos4 x + sin4 x.

c) C = cos12x .

Bit rng cos4x =1

3.


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2.2. Cc phng trnh lng gic c bn Chng 2. Phng trnh lng gic

2.2 Cc phng trnh lng gic c bn

2.2.1 Phng trnh sin x = m

iu kin c nghim: 1 m 1.

Nghim ca phng trnh l : x = arcsin m + k2x = arcsin m + k2 (k Z)

Ch : cc gi tr c bit ca m l m = 0, 1, 12

,

2

2,

3

2

Phng trnh c bit:

sin x = 1 x = 2

+ k2

sin x = 1 x = 2

+ k2

sin x = 0 x = k

(k Z)

Dng sin A = sin B

A = B + k2

A = B + k2

2.2.2 Phng trnh cos x = m

iu kin c nghim: 1 m 1.

Nghim ca phng trnh l :

x = arccos m + k2

x =

arccos m + k2

(k Z)

Phng trnh c bit:cos x = 1 x = + k2cos x = 1 x = k2cos x = 0 x =

2+ k

(k Z)

Dng cos A = cos B

A = B + k2

A = B + k2

2.2.3 Phng trnh tan x = m, cot x = m iu kin c nghim m R Nghim tan x = m x = arctan m + k

cot x = m x = arccot m + k Dng tan A = tan B A = B + k

cot A = cot B A = B + k Ch khi gii phng trnh cha tan, cot ta phi t iu kin cho bin.


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2.2.4 Cc v d


a) sin x = 12.

b) sin

3x 4

= cos2x +

6

.

c) cos

5x

3

= cos2x.

Gii: a)

x = 6

+ k2

x =7

6

+ k2

(k Z)

b) pt sin

3x 4

= sin

2 2x

6

sin

3x

4

= sin

2x +

3

3x

4= 2x +

3+ k2

3x 4

= 2x 3

+ k2

x =

7

12+ k2

x =11

60+

k2

5

(k Z).

c) pt cos5x 3

= cos ( 2x) 5x 3 = 2x + k2

5x 3

= 2x + k2

7x =4

3+ k2

3x = 23

+ k2

x =4

21+ k

2

7

x = 29

+ k2

3

(k Z)

V d 2.2: Gii cc phng trnh:

a) tan x. cot2x 2tan x + 3cot2x = 23

b) tan3x = cot

x +2

5

.

c) tan x +2

cot x= 3.

d) sin2x cos x + 2 sin x 1 = 0.

e) sin2x 3cos x 2sin x + 3 = 0.


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Gii: a) pt tan x + 3 . (cot2x 2) = 0 tan x = 3cot2x = 2

x =

3+ k

2x = arccot 2 + k

x =

3+ k

x =1

2

arccot 2 + k

2

b) pt tan3x = tan

2 x 2

5

tan3x = tan

10 x

3x = 10

x + k x = 40

+ k

4

c) pt tan x + 2 tan x = 3 tan x = 1 x = 4

+ k

2.2.5 Bi tp


a) sin(3x + 1) =

3

2

b) cos(2x + 2) = 12

c) sin2x = sin( x2+ 1)


a) cos2x = sin 5x

b) sinx

2= cos4x

c) cos( sin x) = cos(3 sin x)

d) 2sin x cos x = cos x 2sin x + 1Bi 2.7: Gii cc phng trnh sau:

a) tan x = 1b) tan(2x 1) = tan(x + 2)c) cot(x + 1) = tan2x

d) tan x =

2

cot x

+ 1


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2.3. Cc phng trnh lng gic khc Chng 2. Phng trnh lng gic

2.3 Cc phng trnh lng gic khc

2.3.1 Phng trnh a sin x + b cos x = c

Ta c a sin x + b cos x = c aa2 + b2

sin x +b

a2 + b2cos x =

ca2 + b2

.

Gi l gc sao cho cos =

aa2 + b2

sin =b

a2 + b2

Khi phng trnh tng ng vi:

cos sin x + sin cos x =c

a2 + b2 sin(x + ) = c

a2 + b2

iu kin c nghim:

ca2 + b2 1 |c| a2 + b2 a2 + b2 c2


a) sin x =

3cos x + 1.

b) cos2x 3sin x cos x = 3.

c) sin x cos x =

2cos3x.d) 3sin3x + 4 cos 3x = 5 sin 4x

Gii: a) pt sin x 3cos x = 1 2sin

x 3

= 1 sin

x

3

=

1

2= sin

6

x 3

=

6+ k2

x 3

=5

6+ k2

x =

2+ k2

x =7

6+ k2

(k Z)

b) pt cos2x 32

sin2x = 3.

V a2 + b2 =13

4< c2 = 9 nn phng trnh v nghim.

c) pt 2sin

x 4

=

2cos3x sin

x

4

= sin

2 3x

x

4=

2 3x + k2

x 4

= 2

+ 3x + k2 x =

3

16+ k

2

x = 38

k (k Z

)


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d) pt 35

sin3x +4

5cos3x = sin 4x

Gi l gc sao cho cos =3

5v sin =

4

5ta c:

cos sin3x + sin cos3x = sin 4x sin(3x + ) = sin4x

3x + = 4x + k23x + = 4x + k2

x = k2x =

+ k27

(k Z)

2.3.2 Phng trnh ng cp cha sin v cos

Nhn bit: phng trnh cha sin x, cos x tha mn bc ca tt c cc hng tu l s chn hoc l s l. Chng hn :

- sin x, cos x bc 1.sin2 x, cos2 x, sin x cos x, cos2x, sin2x bc 2.sin3 x, sin2 x cos x, sin x cos2 x, cos3 x, sin3x, cos3x u c bc 3.

- Phng trnh a sin x + b cos x = c c th c coi l phng trnh ng cp

theo sinx

2, cos

x

2.

Cch gii: Ta xt 2 trng hp sau:

- Trng hp 1: cos x = 0.

- Trng hp 2: cos x= 0. Khi ta s chia c 2 v cho cosm x ( m l bc

ca phng trnh ng cp)

V d 2.4: Gii phng trnh 9sin x + 5 cos x 20cos3 x 16sin5 x = 0Gii: Phng trnh tng ng vi

9sin x(sin2 x + cos2 x)2 +5cos x(sin2 x + cos2 x)2 20cos3x(sin2 x +cos2 x) 16sin5x = 0+ TH 1: Nu cos x = 0: Phng trnh tr thnh 9sin x 16sin5x = 0 (V nghim)+ TH 2: Nu cos x = 0: Chia c 2 v cho cos5 x ta c:9t(1 + t2)

2+ 5(1 + t2)

2 20 (1 + t2) 16t5 = 0 t = tan x, phng trnh bc 5, gii phng trnh ta c nghim.V d 2.5: Gii phng trnh:

a) 1 + 2 sin 2x = 6 cos2 x.

b) 2sin3 x = cos xGii: a) pt sin2 x + cos2 x + 4 sin x cos x = 6 cos2 x

sin2 x + 4 sin x cos x 5cos2 x = 0.+) Nu cos x = 0 th ta c sin2 x = 0 sin x = 0. iu ny mu thun.

+) Nu cos x = 0 chia c 2 v cho cos2 x = 0 ta c:

tan2 x + 4 tan x 5 = 0

tan x = 1

tan x = 5

x = 4 + kx = arctan(5) + k

(k Z)


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b) pt 2sin3 x = cos x(sin2 x + cos2 x) 2sin3 x + sin2 x cos x + cos3 x = 0.+) Nu cos x = 0 th 2sin3 x = 0 sin x = 0. iu ny mu thun.+) Nu cos x = 0. Chia c 2 v cho cos3 x.pt 2tan3 x + tan2 x + 1 = 0 (tan x + 1)(2 tan2 x tan x + 1) = 0

tan x = 12tan2 x tan x + 1 = 0( V nghim ) x = 4 + k (k Z).

2.3.3 i s ha phng trnh lng gic

Nhn dng: phng trnh c th qui v phng trnh bc 2, 3, 4 theo sin, cos, tan,cot l nhng phng trnh c th i s ha c. Hoc phng trnh khi dngphp t n ph cng qui v c phng trnh i s.

Phng php: s dng cc cng thc gc nhn i, nhn ba, v cng thc tnhsin x, cos x, tan x, cot x theo t = tan

x

2, ...

V d 2.6: Gii cc phng trnh lng gic sau:

a) cos2x sin x + 1 = 0.

b) sin x + tanx

2= 2

c) cos3x + cos 2x cos x 1 = 0d) tan x + tan2 x + tan3 x + cot x + cot2 x + cot3 x = 6

e) cos3x + 4cos 2x + 4 cos x + 1 = 0.

f) tan x + 2 sin 2x cos2x = 1Gii: a) t t = sin x qui v phng trnh bc 2 n t.

b) t t = tanx

2qui v phng trnh bc 4 n t

c) t t = cos x qui v phng trnh bc 3 n t.d) t t = tan x + cot x qui v phng trnh bc 3 n t

2.3.4 Phng trnh i xng sin, cos

Nhn dng: phng trnh c dng : f(sin x, cos x) = 0 hoc f(sin x, cos x) = 0 f(x, y) l mt hm i xng theo x, y hay ni cch khc f(x, y) = f(y, x).

Cch gii: t t = sin x + cos x = 2sin(x +

4)

hoc t t = sin x cos x = 2sin(x 4

)

vi 2 t 2


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V d 2.7: Gii phng trnh sau:

a) sin x cos x + 2 sin 2x = 1b) sin3 x + cos3 x = 2(sin x + cos x) 1Gii: a) t t = sin x

cos x v gii phng trnh bc 2 n t.

b) t t = sin x + cos x v gii phng trnh bc 3 n t.

2.3.5 Phn tch thnh nhn t

Nhn dng: Bi ton dng ny rt kh nhn dng thng thng theo trc gic cangi gii ton l chnh. Phng php c th dng gii hoc lm cho bi tontr nn n gin hn.

Di y l mt s h c tha s chung hay gp:

f(x) Biu thc cha tha s f(x)

sin x sin2x, sin3x, tan x, tan2x, tan3x,

cos x sin2x, cos3x, tan2x, cot3x, cot x,

1 + cos x cos2

x

2, cot2

x

2, sin2 x, tan2 x

1 cos x sin2 x2

, tan2x

2, sin2 x, tan2 x

1 + sin x cos2 x, cot2 x, cos2

4 x

2

, sin2

4+

x

2

1 sin x cos2 x, cot2 x, cos24 + x2 , sin24 x2sin x + cos x cos2x, cot2x, 1 + sin 2x, 1 + tan x, 1 + cot x, tan x cot x

cos x sin x cos2x, cot2x, 1 sin2x, 1 tan x, 1 cot x, tan x cot x


a) sin2(x ) sin(3x ) = sin x


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b) cos3x 2cos2x + cos x = 0

c) sin2 x + sin2 2x + sin2 3x =3

2

d) cos3x cos4x + sin 2x sin5x =1

2

(cos2x + cos 4x)

Gii: a) pt sin2x + sin 3x = sin x 2sin x cos x + 2 cos 2x sin x = 0 2sin x(cos x + cos 2x) = 0

b) pt 2cos2x cos x 2cos2x = 0 2cos2x(cos x 1) = 0c) S dng cng thc h bc

d) pt 12

(cos7x + cos x) 12

(cos7x cos3x) = 12

(cos2x + cos 4x)

cos x + cos 3x = cos 2x + cos 4x

2cos2x cos x 2cos3x cos x = 0 2cos x(cos2x cos3x) = 0

2.3.6 S dng bt ng thc

Phng php: i vi dng A = 0.Ta chng minh A 0, x t suy ra du bng trong bt ng thc phi xy ra.Gii nghim t iu kin xy ra du bng .i vi dng bin th: A = B.

Ta chng minh : A v B , x dn n phng trnh tng ng viA = = B v du bng 2 bt ng thc xy ra.


a) sin(x +

4) + 2 sin 2x 3 = 0.

b) sin4 x + cos4 x = 2 cos7 x

Gii: a) sin(x +

4

) + 2 sin 2x

3

0 do phng trnh tng ng vi: sin(x + 4) = 1sin2x = 1

b) Ta c sin4 x + cos4 x sin2 x + cos2 x = 1v 2 cos7 x 1. Do m ta c h:

sin4 x = sin2 x

cos

4

x = cos

2

xcos7 x = 1 cos x = 1

x = k2


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2.3.7 Loi nghim khng thch hp

t vn : Trong khi gii mt phng trnh lng gic ta gp bi ton m nghimtm thy khng nm trong tp xc nh. Thiu st trn thng gp khi ta "qun"t iu kin hoc t iu kin nhng khi gii xong khng i chiu vi iu kindn n tnh trng "tha nghim".

Phng php: c rt nhiu phng php nh gii phng trnh nghim nguyn,th nghim trc tip vo iu kin, ... nhng vi mc thi i hc hin nayth phng php n gin d dng nht l biu din tp nghim trn ng trnlng gic. Nhng im b trng vi im b loi tp xc nh s b loi.

V d 2.10: Gii phng trnh sau:

a)4sin6 x + 4 cos6 x 1

2sin x 2 = 0

b) 9 x2 (2 sin x 1) = 0

Gii: a) iu kin: sin x =

2

2.

pt 4sin6 x + 4 cos6 x 1 = 0

4

1 + cos 2x

2

3+ 4

1 cos2x

2

3 1 = 0

1 + 3 cos2 2x 1 = 0 cos2x = 0

x =

4

+ k

2

.

Biu din trn ng trn lng gic ta c nghim:

x = 4

+ k2, x =5

4+ k2

b) Nghim x = 3.Trng hp 3 < x < 3: pt sin x = 1

2Kt hp iu kin ta c nghim:

x =

6, x =

5

6

2.3.8 Bi tp

Phng trnh dng a sin x + b cos x = c.

Bi 2.8: Gii phng trnh sau:

a) cosx

2+

3sin

x

2= 1

b) 3cos x + 4 sin x + 63cos x + 4 sin x + 1

= 6

c) sin3x 3cos3x = 2 sin 2x


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d) 3sin3x 3cos9x = 1 + 4 sin3 3xe) sin x + cos x =

2sin5x

f) cos7x sin5x = 3(cos 5x sin7x)Bi 2.9: Tm gi tr ln nht, nh nht ca hm s sau:

a) y =sin x + 2 cos x + 1

sin x + cos x + 2.

b) y =sin x

2 + cos x

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Phng trnh ng cp:

Bi 2.10: Gii phng trnh sau:a) 4cos2 x 6sin2 x + 5 sin 2x 4 = 0b) 2sin3 x = cos x

c) 6sin x 2cos3 x = 5 sin 2x cos x

d) sin2 x 2cos2 x = 12

sin2x

e)

14cos4 x + 16 cos3 x sin x + 6 sin2 x cos2 x

cos x sin3 x + sin4 x = 2

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

i s ha phng trnh lng gic:


a) cos2x + sin2 x + 2 cos x + 1 = 0

b) 2cos3x 7cos2x + 7 cos x 2 = 0.

c) cos4 x + sin4 x = 2 cos(2x + 4

)cos(2x 4

)

d) sin4 x + cos4 x cos2x + sin2 2x

4 2 = 0

e) tan2 x 4cos x

+ 5 = 0

f)3

sin2 x+ 3 tan2 x 11(tan x + cot x) 1 = 0

g) sin4 x + (1 + sin x)4 = 17

h) 2cos26x

5+ 1 = 3 cos

8x

5


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6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Phng trnh i xng:


a) sin3 x + cos3 x = 1 sin2x2b) sin2x + 4(cos x sin x) = 4

c) cos x +1

cos x+ sin x +

1

sin x=

10

3

d) sin x cos x = 6(sin x + cos x 1)e) 1 + sin 2x = sin x + cos x

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Phn tch thnh nhn t:


a) sin x + cos x + 1 = 2 cos(x2

4

)

b)

2(cos4 x sin4 x) = sin x + cos xc) sin x + 3 sin 2x = sin 3x

d) cos x sin2x cos3x =sin4x

4

e) cos2x + sin 2x + cos x + 3 sin x + 1 = 0

f) (2 sin x 1)(2sin2x + 1) = 3 4cos2 xg) cot x tan x = sin x + cos x

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

S dng bt ng thc:


a) sin3 x + cos3 x = 2 sin4 xb) cos3x +

2 cos2 3x = 2(1 + sin2 2x)

c) cos(x) = x2 4x + 5

d) 2sin(x +

4) = tan x + cot x

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Loi nghim khng thch hp


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a)sin6 x + cos6 x

tan(x 4

)tan(x + 4

)= 1

4

b)sin6 x + cos6 x

cos2 x sin2 x=

tan2x

4

c)sin x (sin x + cos x) 1

cos2 x + sin x + 1= 0

d) tan x (2sin x3

1) = 0

e)1 + tan x

1 tan x = 1 + sin 2x

f) tan2 x = 1 + cos x1 sin x

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Dng ton tng hp:


a) sin x + cos 2x + cos2x (tan2x 1) + 2sin3x = 0.

b) cos x. cos2x. cos4x. cos8x = 116

c)sin4x + cos4x

5sin2x=

1

2cot2x 1

8sin2x

d) 2sin2x + sin 2x + 3 sin x cos x + 1 = 0

e) 4sin2x

2 3cos2x = 1 + 2cos2

x 3

4

f) 22cos3x 4 3cos x sin x = 0

g) tan

2+ x

3tan2x = cos2x 1

cos2x

h) tan

3

2 x

+sin x

1 + cos x= 2

i) cos23x. cos2x

cos2x = 0

j) 2sin x (1 + cos 2x) + sin 2x = 1 + 2 cos x

k) sin3x 3cos3x = sin xcos2x 3sin2x cos x


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l)1

sin x+

1

sin

x 3

2

= 4 sin74

x

m)

2cosx 7

4 = cos2x 7

2 +1

2

n) sin2x + sin x 12sin x

1sin2x

= 2cot 2x

o) sin

5x

2

4

cos

x

2

4

=

2cos

3x

2

p) 2

2sin

x

12

cos x = 1.

q) 2cos2x + 23sin x cos x + 1 = 3 sin x + 3cos xr)

sin2x

cos x+

cos2x

sin x= tan x cot x


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Chng 3. Phng trnh cha cn v du gi tr tuyt i

Chng 3Phng trnh cha cn v du gitr tuyt i

3.1 Phng trnh cha du gi tr tuyt i3.1.1 Kin thc cn nh

Hc sinh cn nh mt s tnh cht sau ca du gi tr tuyt i:

|A| =

A Nu A 0A Nu A < 0

| A| = |A|

|A2

|= A2

|A| A. Du ng thc khi A 0.|A| A. Du ng thc khi A 0

|A| + |B| |A + B|.Du ng thc xy ra A.B 0.

|A| + |B| |A B|.Du ng thc xy ra A.B 0.

3.1.2 Cc dng bi tp Nguyn tc chung: gii mt phng trnh cha du gi tr tuyt i, ta c

th bm vo nh ngha xt du ca biu thc trong du gi tr tuyt i, phntrng hp ph du gi tr tuyt i. Tuy nhin, nhiu khi vic xt du mtbiu thc l kh kh khn, di dng nn y ti a ra mt s cch bin itng i vi mt s dng phng trnh cha du gi tr tuyt i c bn hcsinh c th p dng n lm cc bi tp mt cch chnh xc ngn gn.

- Dng 1

|A| = B A 0A = B A < 0

A = B A = BA = B

B 0


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3.1. Phng trnh cha du gi tr tuyt i Chng 3. Phng trnh cha cn v du gi tr tuyt i

- Dng 2:|A| = |B| A2 = B2 A = B

- Dng 3: t n ph vi biu thc cha du gi tr tuyt i.

- Dng 4: S dng bt ng thc nh gi 2 v ca phng trnh :

Chng hn nu ta c phng trnh A = B. Ta chng minh:A v B t ta c A = = B.

- Dng 5 : Lp bng xt du vi biu thc trong du gi tr tuyt i ph dugi tr tuyt i.

3.1.3 Cc v d


a) |x2

+ x 1| = 2 x ()b) |2x2 + 3x 2| = |x2 x 3| ()

Gii: a) ()

2 x 0 (1)x2 + x 1 = 2 x (2)x2 + x 1 = x 2 (3)

(1) x 2(2) x2 + 2x 3 = 0

x = 1

x = 3tha mn iu kin

(3)

x2 + 1 = 0 Phng trnh v nghim

b) ()

2x2 + 3x 2 = x2 x 3 (1)2x2 + 3x 2 = x2 + x + 3 (2)

(1) x2 + 4x + 1 = 0 x = 2 3

(2) 3x2 + 2x 5 = 0 x = 1

x = 53

V d 3.2: Gii phng trnh x4 + 4x2 + 2 |x2 2x| = 4x3 + 3Gii: Phng trnh (x2 2x)2 + 2 |x2 2x| 3 = 0t t = |x2 2x| 0. Phng trnh tr thnh:

t2 + 2t 3 = 0

t = 1tha mn

t = 3 (loi)

+)t = 1 x2 2x = 1 x2 2x = 1x2 2x = 1

x = 1 2x = 1

V d 3.3: Gii phng trnh |x + 1| + |2 x| = x4 + 4x2 1


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3.2. Phng trnh cha cn thc Chng 3. Phng trnh cha cn v du gi tr tuyt i

Gii: Ta c V T = |x + 1| + |2 x| |x + 1 + 2 x| = 3 x (1)m V P = x4 + 4x2 1 = (x2 2)2 + 3 3 x (2).T (1) , (2) V T 3 V P xDo phng trnh cho

(x + 1) . (2 x) 0x2 2 = 0 x =

2

V d 3.4: Gii phng trnh 2 |x + 1| |x2 2x 8| = 5 x + x2

Gii: Ta c bng xt du:

x 2 1 4 +x + 1 0 + +

x2 2x 8 + 0 0 +

Trng hp 1: x < 2.Phng trnh

2 (

x

1)

(x2

2x

8) =

5

x + x2

2x2 x 11 = 0 x = 1 + 894 (loi)

x =1 89

4(tha mn)

Trng hp 2: 2 x < 1Phng trnh 2 (x 1) + (x2 2x 8) = 5 x + x2

3x = 5 x = 53

(tha mn)

Trng hp 3:

1

x < 4

Phng trnh 2 (x + 1) + (x2 2x 8) = 5 x + x2 x = 1 (tha mn)

Trng hp 4: x 4Phng trnh 2 (x + 1) (x2 2x 8) = 5 x + x2

2x2 5x 15 = 0

x =5 +

145

4(tha mn)

x =5 145

4(loi)

Vy phng trnh c tp nghim S = 1 894

; 53

; 1;5 + 145

4

3.2 Phng trnh cha cn thc

3.2.1 Cc dng bi tp

A = B

A = B2

B 0

A =

B A = B

A 0 A = B

B 0


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3.2. Phng trnh cha cn thc Chng 3. Phng trnh cha cn v du gi tr tuyt i

t n ph. Ta t cn thc hoc biu thc trong cn bng mt n ph lmphng trnh tr nn n gin hn.

S dng bt ng thc. Cch lm tng t phn phng trnh cha du gi trtuyt i.

3.2.2 Cc v dV d 3.5: Gii phng trnh sau:

a)

x + 3 = 2 xb)

x2 + x 1 = 2x + 3

Gii: a) Phng trnh

x + 3 = (2 x)22 x 0

x2 4x + 4 = x + 3x 2

x2

5x + 1 = 0

x 2 x =5 21

2x 2 x =5

21

2

b) Phng trnh

x2 + x 1 = 2x + 32x + 3 0

x2 x 4 = 0

x 32

x =1 17

2

x 32

x = 1 +

17

2

V d 3.6: Gii phng trnh sau: 2x2 2x 2 = x2 + 2x + 5.Gii: t t =

x2 2x 2 0 x2 2x = t2 + 2.

Phng trnh 2t = t2 2 + 5 t2 + 2t 3 = 0

t = 1

t = 3 (loi)Ta c t = 1 x2 2x 2 = 1 x2 2x 3 = 0

x = 1x = 3

Bnh lun: Nu phng trnh ny ta gii theo cch ca phng trnh cn thcc bn th s gp mt phng trnh bc 4. Khi mi vic tr nn phc tp hn

nhiu.V d 3.7: Gii phng trnh :

x + 1 +

3 x = 2 (x2 2x + 3)

Gii: p dng bt ng thc Bunhiacopxki vi 2 bx + 1;

3 x

, (1; 1)

ta c: V T2 =

x + 1.1 +

3 x.12 x + 12 + 3 x2 . (12 + 12). V T2 (x + 1 + 3 x) .2 = 8 V T 22 (1)Du bng xy ra khi

x + 1

1=

3 x

1 x = 1

V P =

2 (x2 2x + 3) = 2. (x 1)2 + 2 22 (2)Du bng xy ra khi x = 1.T (1), (2) ta c phng trnh cho c duy nht nghim x = 1.


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3.3. Bt phng trnh cha du gi tr tuyt i Chng 3. Phng trnh cha cn v du gi tr tuyt i

3.3 Bt phng trnh cha du gi tr tuyt i

3.3.1 Dng c bn

|A| < B B < A < B

|A| > B A > BA < B

|A| < |B| A2 B2 < 0 (A B) . (A + B) < 0

3.3.2 Cc v d

V d 3.8: Gii cc bt phng trnh sau:

a) |x2 2x 1| < x + 1

b) |2x2 + x 3| 2x + 1c) |x2 + x 1| < |2x2 + x 2|

Gii: a) Bt phng trnh

x2 2x 1 < x + 1x2 2x 1 > x 1

x2 3x 2 < 0x2 x > 0

3

17

2< x 1 x < 0

x

3 172

; 0

1;3 +

17

2

b) Bt phng trnh 2x2 + x 3 2x + 12x2 + x 3 2x 1 2x2 x 4 02x2 + 3x 2 0

x 1 +

33

4 x 1

33

4

2 x 12

x 12

x 1 +

33

4

c) Bt phng trnh (2x2 + x 2)2 > (x2 + x 1)2 (2x2 + x 2)2 (x2 + x 1)2 > 0 (x2 1)(3x2 + 2x 3) > 0

(x 1) (x + 1)x 1

10

3 x 1 +

10

3 > 0 x < 1

10

3 1 < x < 1 +

10

3 x > 1

3.4 Bt phng trnh cha cn thc

3.4.1 Dng c bn

A > B A > B2

B 0B < 0

A 0


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3.4. Bt phng trnh cha cn thc Chng 3. Phng trnh cha cn v du gi tr tuyt i

A < B

A < B2

A 0B 0

A x + 2

b) 2x2 + 3x 2 x + 1c)

x2 + 3x + 2 (x + 2)2x + 2 0x + 2 < 0

x2 + x 1 0

3x + 5 < 0

x 2

x < 2x 1 +

5

2 x 1

5

2

2 x < 53x < 2

x < 53

b) Bt phng trnh

x + 1 02x2 + 3x 2 02x2 + 3x 2 (x + 1)2

x 1x 2 x 1

22x2 + 3x 2 x2 + 2x + 1

x 12x2 + x 3 0

x

1

21 132

x 1 +13

2

12

x 1 +

13

2

c) Bt phng trnh

x2 + 3x + 2 0x2 + 3x + 2 < 3x2 7x + 4

x 1 x 22x

2

10x + 2 > 0 x 1 x 2

x >5 +

21

2 x 5 +

21

2


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3.5. Bi tp Chng 3. Phng trnh cha cn v du gi tr tuyt i

3.5 Bi tp

Phng trnh cha du gi tr tuyt i:


a) |2x 1| + |2x + 1| = 4b) |x2 3x + 2| 2x = 1c) |x2 + x 12| = x2 x 2d) |2 |2 x|| = 1e) |x2 2x| = |2x2 1|f) |x2 5x + 4| 9x2 5x + 9 + 10x |x| = 0

Hng dn. a) x = 1; 1

b) x =5 21

2

c) x = 5; 7d) x = 5; 3;1;1.

e) x = 1 2; 13

; 1.

f) x = 5 259

18.

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Bt phng trnh cha du gi tr tuyt i:

Bi 3.2: Gii cc bt phng trnh:

a)

|x

3

|+

|5

x

|< 3x

b) |x2 x 6| < xc) |x2 5x + 4| > x 2d) |2x |x 1|| < 2e) |3x2 2x 1| < |x2 x|

Hng dn. a) x

8

5; +

.

b) x 6;1 + 7.c) x ; 2 + 2 3 + 3; +. Bin tp : Th.s Minh Tun Trang 55 Khoa T nhin - Trng CSP Nam nh

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d) x

13

; 1

.

e) x

12

;1

4

.

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Phng trnh cha cn thc:


a) x 2x + 3 = 0b)

x + 4 1 x = 1 2x

c) (x 3) (x + 1) + 4 (x 3) .x + 1x 3 = 3d)

3 + x +

6 x (3 + x) . (6 x) = 3

e) x3 + 1 = 2 3

2x 1Hng dn. a) x = 3.

b) x = 0.

c) x = 1

5; 1

13.

d) x = 3; 6.

e) x = 1; 12

5.


a)

x 2 + 4 x = x2 6x + 11b) 3

x + 34

3

x

3 = 1

c)

x + 2

x 1 +

x 2x 1 = x + 32

d)x2

3x 2

3x 2 = 1 x

e) x2 +

x + 5 = 5

f) 3

x 1 + 3x 2 = 32x 3

Hng dn. a) x = 3b) x = 61; 30.c) x = 1; 5.


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d) x = 1.

e) x = 1 + 17

2;

1 212

.

f) x = 1; 2.

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

Bt phng trnh cha cn thc :


a)

x + 9 +

2x + 4 > 5

b) 4(x + 1)2 < (2x + 10)

1 3 + 2x2c) (x

3)

x2 + 4

x2

9

d)

x + 2 x + 1 xe)

2x2 6x + 1 x 2

f)x

x + 1 2

x + 1

x> 3

g)

5x2 + 1

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