Linearne jednačine

Def. Dve jednačine su ekvivalentne ukoliko su sva rešenja prve istovremeno rešenja druge i obrnuto.

Primer:

1 x(x+5)=0

2 x+5 =0

Da li su jednačine 1 i 2 ekvivalentne?

1 x(x+5)=0 x =0 x= -5

2 x+5 =0 x= -5

Nisu ekvivalentne jer rešenje 1 x=0 nije i rešenje jednačine 2

1. Napiši ekvivalentnu konjukciju datoj:

x+3 0 x2 – 9 =0 x -3 (x+3)(x- 3)=0 x -3 (x+3=0 x- 3=0) x -3 (x= -3 x=3) (x -3 x= -3) ( x -3 x =3) x x= 3 x= 3

2. 5x-6 = 0 25x2−360

x = 65 (5x-6)(5x+6) 0

x = 65 (5x-6 0 5x+6 0)

( x = 65 5x-6 0) ( x =

65 5x+6 0)

( x = 65 x

65 ) ( x =

65 x

−65 )

x x= 65

3. Da li su sledeće jednačine ekvivalentne:x+2=2x-3 i (x-5)(x+2)= (x-5)(2x-1) ?

x+2=2x-3 x-2x= -3-2 -x= -5 x= 5

(x-5)(x+2)= (x-5)(2x-1) (x-5)(x+2)- (x-5)(2x-1) = 0 (x-5) (x+2 -2x+1)=0 (x-5) (3-x)=0 - (x-5) (x-3)=0 / * (-1) (x-5) (x-3)=0 x-5=0 x-3=0 x=5 x=3Nisu ekvivalentne jer nemaju ista rešenja

4. Zašto jednačina x+8= 4x-3(x-223 ) ima beskonačno mnogo rešenja ?

x+8= 4x-3(x-223 )

x+8 = 4x-3 (x- 83 )

x+8 = 4x – 3* 3x−83

x+8 = 4x – (3x-8)

x+8 = 4x – 3x+ 8

x+8 = x+ 8

jednačina je neodređena pa zato ima beskonačno mnogo rešenja.

5. Rešiti po y jednačine:a) 1,3(y - 0,7)- 0,12 (y+10) = 5y- 9,75

1,3y – 1,3* 0,7 - 0,12 y- 0,12* 10 = 5y- 9,75

1,3y – 0,91 - 0,12 y- 1,2 = 5y- 9,751,3y – 0,91 - 0,12 y- 1,2 = 5y- 9,751,18 y – 2,11 = 5y- 9,75 1,18 y -5y= - 9,75+2,11-3,82y = - 7,64 y= 2

b) 5 (y+2)(2y+3) - 2(y-1)(5y-4) =75

10 y2 +15 y+20y +30 - 2( 5y2 -4y-5y+4) =7510 y2 +35 y +30 - 2( 5y2 -9y+4) =7510 y2 +35 y +30 - 10y2 + 18y - 8 =7553y +22 =7553y= 53y=1

c) (6y-3)(2y+1) – 5 (2y+1)2+ (3y-1)2 = (y-1)2

12y2+6y-6y-3 -5 *(4y2+4y+1)+ 9y2-6y+1= y2-2y+112y 2 - 3 -20y 2 -20y - 5+ 9y 2 -6y+1= y2-2y+1y2 – 7 - 26y = y2-2y+1y2 – 7 - 26y - y2+2y – 1 = 0-24y -8 =0-24y=8

y= −13

6. Rešiti jednačine:

a)2−3 x8

+x−4 x+34 = −2

12

2−3 x8

+x−4 x+34 = −52 /*8

2-3x+8x – 2(4x+3)= -5*42+5x – 8x – 6 = -202 – 3x – 6 = -20– 3x – 4 = -20 – 3x = -20+4 – 3x = - 16

x= 163

b)x+25

−3= x−12

−x /*10

2(x+2)-30 = 5(x-1)-10x2x+4-30= 5x-5-10x2x-26 = - 5x-52x + 5x = -5+267x = 21 x= 3

7. a) 1+ x42

+ 7 x2+1

6−1+5 x

24−

72+6x

12=13

/*24

12*(1+x4 )+ 4*(

7 x2

+1) – (1+5x) – 2*(72+6x) = 8

12+3 x+ 14x+4 – 1-5x – 7−12 x = 812+17x+3−¿5x – 7−12 x = 8 15+12 x – 7−12 x = 8 15−¿7 = 8 8=8jednačina je neodređena tj. ima beskonačno mnogo rešenja

b) −12x+2 x−10−7 x

32

+x−1+x

33

=1 /*6

-3x+ 3*(2 x−10−7 x3 )+2* (x−

1+x3 ) =6

-3x+ 6x−¿)+2x−2∗1+x3 = 6

3x−10+7 x+ 2x−2∗1+x3 = 6

12x−10−2∗1+x3 = 6 /*3

36x−30−2∗(1+x) = 18

36x−30−2−2 x = 18

34x−32 = 18 34x = 18+32 34x= 50 / :2 17 x= 25

x= 2517

8. a) 1,3−3 x2

−1,8−8 x1,2

=0,4−5 x3 /*

1010

13−30 x20

−18−80 x12

= 4−50x30 /*2

13−30 x10

−18−80 x6

= 4−50x15 /*3

39−90 x10

−18−80 x2

=4−50 x5 /*10

39-90x- 5*(18-80x) = 2*(4-50x)

39-90x- 90+400x = 8-100x

39-90x- 90+400x = 8-100x

-51+310x = 8-100x

410x = 59

9. x2−4x+2

=0 ( x−2 ) (x+2 )=0 x+20

(x=2 x= -2) x−2

(x=2 x−2¿(x−2 x= -2)

x=2 x x=2

AB=0 A=0B0

10. x−3x2−9

=0 x−3=0 x2−90 x−3=0 (x−3 ) (x+3 )0

x=3( x3 x−3) (x=3 x3) x -3 x x -3 x

11. x−3x+3

+ 3 x−13 x+1

=2 ( x−3 ) (3x+1 )+(3 x−1 )(x+3)

( x+3 )(3 x+1) = 2 3x2+x−9 x−3+3x2+9 x−x−3

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

6 x2−6( x+3 )(3 x+1)

=2 6 x2−6( x+3 )(3 x+1)

=2¿( x+3 )(3x+1)( x+3 )(3x+1)

6 x2−6( x+3 )(3 x+1)

- 2¿( x+3 )(3x+1)( x+3 )(3x+1)

= 0

6 x2−6( x+3 )(3 x+1)

−2∗3 x2+ x+9x+3

( x+3 )(3 x+1)=0 6 x

2−6−6 x2−20 x−6( x+3 )(3 x+1)

=0

−20 x−12( x+3 )(3 x+1)

=0

−20 x−12=0 ( x+3 ) (3x+1 ) 0

x= −35 (x -3 x

−13 )

x= −35

12. 2924

− 4x−8

= 32x−16

− 23 x−24

2924

− 4x−8

= 32(x−8)

− 23(x−8) /*6

294− 24x−8

= 9x−8

− 4x−8

AB = 0 (A=0 B =0)

AB 0 (A 0 B 0)

294= 24x−8

+ 9x−8

− 4x−8

294= 29x−8 / :29

14= 1x−8

1x−8

−14=0 4−x+8

4(x−8)=0

12−x4 (x−8)

=0 x=12 x 8

13. y+12 y−1

− 11 y+512(2 y−1)

= y−34−8 y

+ 16

y+12 y−1

− 11 y+512(2 y−1)

= 3− y4 (2 y−1)

+ 16 /*12

12( y+1)2 y−1

−11 y+52 y−1

=9−3 y2 y−1

+2

12 y+12−11 y−52 y−1

=9−3 y2 y−1

+2 y+72 y−1−9−3 y2 y−1

=2 y+7−9+3 y2 y−1 = 2 4 y−22 y−1

= 2*2 y−12 y−1

4 y−22 y−1 - 2*

2 y−12 y−1 =0 4y-2-4y+2 =0 2y-10 0= 0 y

12 xR

12

14. 1

8 x−16+ 5−x8 x−4 x2

= 78 x

− x−12 x (x−2)

1

8(x−2)− 5−x4 x (x−2)

= 78 x

− x−12x (x−2) /*8

1x−2

− 10−2xx (x−2)

=7x− 4 x−4x (x−2)

x−10+2 x+4 x−4x (x−2) ¿ 7x

7 x−14x(x−2) ¿

7x

7 x−14x(x−2)

−7x = 0

7 x−14−7(x−2)x (x−2) =0

7 x−14−7 ( x−2 )=0 x (x−2 )0

7 x−14−7 x+14=0 ¿ x2)

0=0 ¿ x2)

xR 0,2

15. 2

x2−4− 1x2−4 x+4

= 1x2+5 x+6

x2−4 = (x-2)(x+2)x2−4 x+4 = (x-2)2

x2+5x+6 = x2+2x+3 x+6 = x(x+2)+3(x+2)= (x+2)(x+3)

2 ( x−2 ) ( x+3 )− (x+2 ) ( x+3 )−( x−2 )2

( x−2 )2 ( x+2 )(x+3)=0

2 (x2+3 x−2 x−6 )−(x2+3 x+2 x+6 )−(x2−4 x+4)( x−2 )2 (x+2 )(x+3)

=0

2x2+2x−12−x2−5 x−6−x2+4 x−4( x−2 )2 ( x+2 )(x+3)

=0

x−22( x−2 )2 ( x+2 )(x+3)

=0 x−22=0 ( x−2 )2 ( x+2 ) ( x+3 ) 0

x=22 x2 x -2 x -3 x=22

16. 5x+4x+4

−9−3x−2x2

16−x2+ 13−3 xx−4

=0

5x+4x+4+ 9−3 x−2 x

2

x2−16+13−3 xx−4

=0

(5x+4 ) ( x−4 )+9−3 x−2x2+(13−3x )(x+4 )

( x+4 )(x−4)=0

5x2−20 x+4 x−16+9−3 x−2x2+13 x+52−3 x2−12x(x+4 )(x−4)

=0 −16 x−7+10 x+52−12x

( x+4 )(x−4)=0

−18 x+45( x+4 )( x−4)

=0 x= 52 ( x+4 )(x−4) 0

x= 52 (x 4 x -4) x=

52

17. 1+3 x3x−1

−1−3 x1+3 x

= 121−9 x2

1+3 x3x−1

+ 1−3 x3 x+1

= −129 x2−1

1+6 x+9x2−1+6 x−9x2+12

(3 x−1 )(3 x+1) =0

12 x+12(3x−1 )(3 x+1) =0 12 x+12=0 (3 x−1 )(3 x+1) 0

x= -1 x 13 x

−13 x= -1

18. 2

6 x+1− 31−6 x

= 8+9x36x2−1

2 (6 x−1 )+3 (6 x+1 )−8−9 x

6 x+1 = 0

12x−2+18 x+3−8−9 x6 x+1

= 0

21 x−76 x+1 = 0 x=

13 6x+1 0 x=

13 x

−16

19. x+18−x3

− 12x2−x3

+ 2x4+2 x3+4 x2

+ 1x2

= 0

8−x3=(2−x )(4+2 x+ x2)2 x2−x3=x2(2−x)x4+2 x3+4 x2=x2(x2+2 x+4)

x3+x2−x2−2 x−4+4−2 x+8−x3

x2(x2+2 x+4) (2−x )=0

−4 x+8x2(x2+2 x+4) (2−x )

=0

x=2 x2 (x2+2 x+4 ) (2−x )0 x=2 x 0 x2 x

20. xx−2

−2 x−3x+2

= x2−9 x4−x2

x ( x+2 )−(2 x−3 ) ( x−2 )+x2−9 xx−2

= 0

x2+2 x−(2x2−4 x−3 x+6)+x2−9xx−2

= 0

2x2−7 x−2 x2+7 x−6x−2

= 0 −6x−2 = 0 x-20 -6= 0 x2 jednačina je nemoguća

21. 3

4 y−12− 0,75 y−2y2−6 y+9

= y+2y3−9 y2+27 y−27

4 y−12=4 ( y−3)

y2−6 y+9=( y−3)2

y3−9 y2+27 y−27=( y−3)3

3 ( y2−6 y+9 )−4 (0,75 y−2 ) ( y−3 )−4 ( y+2)4( y−3)3

= 0

3 y2−18 y+27−(3 y−8 ) ( y−3 )−4 y−84 ( y−3)3

= 0

3 y2−22 y+19−(3 y2−9 y−8 y+24)

4( y−3)3 = 0 3 y

2−22 y+19−3 y2+17 y−244 ( y−3)3

= 0

−5 y−54 ( y−3)3

= 0 y= -1 y-30 y=-1 y3 y= -1

22. 4 x

1−6 x+12 x2−8 x3= 12 (1−4 x2)

− 32−8 x+8x2

8 x−1+4 x−4 x2+3−12 x2

2 (1−2 x )3(1+2x )=0

−16 x2+12 x+22 (1−2x )3(1+2 x)

=0 8x2−6 x+1=0 1-2x 0 1+2x 0 x12 x−12

1−6 x+12 x2−8 x3=¿ (1−2x)3 8x2−6 x+1=0

1−4 x2=(1−2 x )(1+2x ) I2 = 8x2 I = √8x= √4∗2 x = 2√2x

2−8x+8 x2=¿ 2(1- 4x+4x2)= 2(1-2x)2 -6x= 2 I*II =2*2√2x∗II II = −32√2 II2=

98

8x2−6 x+1=¿ 8x2−6 x+98−98+1 =

(2√2x− 32√2)2−9

8+1 =

= (2√2x− 32√2)2−1

8=¿ (2√2x− 3

2√2− 12√2)( 2√2x− 3

2√2+ 12√2 )

= (2√2x− 2√2 )( 2√2x− 1

√2 ) /*√2 = (4x -2)(4x - 1)= 2(2x-1)(4x-1)=0 x= 12 x=

14 x=

14

23. Rešiti jednačine po x:

a)3− 1

y

3+ 1y

− 1y=y−13

y+ 13

−13

3 yy− 1y

3 yy

+ 1y

− 1y=

3 y3−13

3 y3

+ 13

−13

3 y−1y

3 y+1y

− 1y=

3 y−13

3 y+13

−13 3 y−13 y+1 −1y = 3 y−13 y+1 −13 −1y

=−13 −1y

+ 13=0

−3+ y3 y

=0 -3+y = 0 3y 0 y = 3 y 0

b)

y+1y−1

−1

y+1y−1

− y−1y+1

=72

y+1y−1

− y−1y−1

( y+1 )( y+1)( y−1 )( y+1)

−( y−1 )( y−1)( y+1 )( y−1)

= 72

y+1− y+1y−1

( y+1 ) ( y+1 )−( y−1 )( y−1)( y−1 )( y+1)

=72

2y−1

( y+1− y+1 ) ( y+1+ y−1 )( y−1 )( y+1)

=72

2y−12∗2 y

( y−1 )( y+1)

=72

2 ( y−1 )( y+1)4 y ( y−1)

=72

( y−1 )( y+1)2 y ( y−1)

−72=0

y+12 y

−72=0

y+1−7 y2 y

=0 -6y+1 = 0 y0

y = −16 y0

c) 1 −1

1− 11−x

=2 1

−11− x1− x

− 11−x

=2 1

−11− x−11−x

=2 1

+1x1− x

=2

1+ 1−xx

=2 xx+ 1−xx

=2 x+1−xx

=2 1x=2

1x−2=0

1x−2 xx=0

1−2 xx

=0 1-2x = 0 x 0 1=2x x 0 x = 12 x 0

Linearne jednačine sa parametrom

ax=b

1) a 0 x = ba

2) a= 0 b 0 tj. 0*x = b 00= b 0

jednačina nema rešenje - nemoguća je

3) a = 0 b= 0 0*x = 0 T x R je rešenje jednačine tj. jednačina je neodređena

1. Rešiti po x jednačine (m je realan parametar):a) m(mx+1)=2(2x-1)

m2 x+m=4 x−2

m2 x−4 x=−m−2

x (m2−4 )=− (m+2 )

x(m-2)(m+2) = - (m+2)

1 tačno jedno rešenje:

(m-2)(m+2) 0 m2 m -2

x= −(m+2)

(m−2 )(m+2)= −1m−2

2 nemoguća:

(m-2)(m+2) = 0 m+2 0 (m=2 m= -2) m -2

(m=2 m -2 ) (m= -2 m -2) m=2 x m=2

za m=2 jednačina nema rešenje.

provera: 0*4x= - 4 0= -4

3 neodređena:

(m-2)(m+2) = 0 m+2 = 0 (m=2 m= -2) m= -2

(m=2 m= -2 ) (m= -2 m= -2) x m= -2 m=-2

provera: -4*0x= -0 0=0

xR je rešenje

b) m3 x−m2−4=4m ( x−1 ) m3 x−m2−4=4mx−4m m3 x−4mx=m2+4−4m mx (m2−4 )=(m−2)2

mx (m−2 )(m+2)=(m−2)2

1 m (m−2 ) (m+2 )0m0m2m−2

x = (m−2)2

m (m−2 )(m+2)= m−2m(m+2)

2 m (m−2 ) (m+2 )=0m2 (m=0m=2m=−2 )m 2 m=0 m= -2

provera: 0*(-2)*2x= (-2)2

0=4

3 m (m−2 ) (m+2 )=0m=2m=2

provera: 2*0*4x = 0

0= 0 T

xR

x= 00

c) m2 x+4=m ( x+4 )m2 x+4=mx+4mm2x−mx=4m−4mx (m−1 )=4 (m−1 )

1 m (m-1) 0 m 0 m 1

x= 4(m−1)m(m−1)

= 4m

2 m (m-1)= 0 m-1 0 (m=0 m= 1) m1 m=0

3 m (m-1)= 0 m-1= 0 (m=0 m= 1) m= 1 m=1

2. Odrediti parametar a tako da jednačina bude nemoguća:

(x+a)2= x(x+a)+4a2 x2+2ax+a2= x2+ax+4a2 2ax+a2= ax+4a2 ax= 3a2 x= 3a2

a

a=0 3a2 0 a=0 a0

jednačina nikad nije nemoguća.

3. Odrediti parametar a tako da jednačina bude neodređena:

a(x −1a2

) + a2 (x−1a )=2 ax –

1a+a2 x−a=2

ax(1+a)= 2 + 1a +a ax(a+1) = a

2+2a+1a

A B

A=0 B=0 (a=0 a= -1) (a= -1 a 0) (a=0 a= -1 a 0) (a= -1 a= -1 a 0)

x a= -1 a= -1

provera: -x*0=0 0=0

xR je rešenje

4. Rešiti jednačine (a,b,c su realni parametri ): (3y-2c)2 – (y-2c)(2y+c) = 7 y2- 12c2

9 y2−12 yc+4 c2−(2 y2+ yc−4 yc−2c2 )=¿7 y2- 12c2

9 y2−12 yc+4 c2−2 y2− yc+4 yc+2c2=¿7 y2- 12c2

7y2 -9yc +6c2 = 7 y2- 12c2

18c2=9 yc

1 c 0

y= 18c2

9c=2c

2 c=0 c20c=0 c 0 jednačina nikad nije nemoguća

3 neodređena: c=0 c2=0 c=0 c = 0 c=0 yR je rešenje

5. (2 y+a )2=4 ( y−1 ) ( y+1 )+4 ( y+1,25a2 )

4y2+4 ay+a2=4 y2−4+4 y+5a2

4ay - 4y = 4a2 -4 4y (a-1)= 4(a2−1)

1 tačno jedno rešenje: a-1 0 a 1

y= 4 (a−1 )(a+1)4(a−1)

=a+1

2 nemoguća:a-1=0 (a 1 a -1) (a=1 a1)(a=1 a -1) x a=1 a=1

3 neodređena:

a=1 (a=1 a= -1) (a=1 a=1) (a=1 a= -1) a=1 a a=1

6. a(b-y)+by+ab= a2+b2

ab-ay+by+ab= a2+b2 -y(a-b)= a2+b2−2ab− y (a−b )=(a−b)2

1 tačno jedno rešenje:

a-b 0 a b

-y = (a−b)2

a−b=a−b y= b-a

2 nemoguća:

a= b (a−b )20a=baba

3 neodređena:

a= b a=b a=b

7. 3ax- (2 x−3a )2= (2 x+a ) (5a−2x )

3ax – 4x2+12ax−9a2=10ax−4 x2+5a2−2ax

7ax = 14a2 ax = 2a2

1 a 0

x= 2a2

a=2a

2 nemoguća: a=0 2a0 a=0 a0 a

3 neodređena: a=0 2a=0 a=0 a=0 a=0

8.(a+b)xa−b −

(a−b )(a−x )a+b =

(2 x+a )(a−b)a+b

(a+b ) (a+b ) x−(a−b ) (a−x ) (a−b )−(2 x+a ) (a−b ) (a−b )

(a−b ) (a+b )=0

a2 x+2abx+b2 x−(a2−2ab+b2 ) (a−x )− (2x+a )(a2−2ab+b2)

(a−b ) (a+b )=0

a2 x+2abx+b2 x−a3+a2 x+2a2b−2abx−ab2+b2x−2a2 x+4 abx−2b2 x−a3+2a2b−ab2

(a−b ) (a+b )=0

−2a3+4 a2b−2ab2+4abx

(a−b ) (a+b )=0

2a(−a2+2ab−b2+2bx )(a−b ) (a+b )

=0

2bx – (a−b)2=0 aba−b x= (a−b)2

2b ab a−b

1 tačno jedno rešenje: 2b 0 b 0 onda je x= (a−b)2

2b

2 nemoguća: b=0 a-b0 b=0 ab b=0

3 neodređena: b=0 b=a a=b=0

9. 3n+xn+3

+ x−3nn−3

=nx−3n2−9

(3n+x ) (n−3 )+( x−3n ) (n+3 )−nx+3

(n+3 )(n−3)=0

3n2−9n+nx−3x++nx+3 x−3n2−9n−nx+3(n+3 )(n−3)

=0

−9n+nx−9n+3

n+3=0 −9n+nx−9n+3=0 n+3 0 n-3 0 nx= 3(6n-1) n+3

0 n-3 0

1 n 0 onda je x= 3(6n−1)

n

2 nemoguća: n=0 6n1 n=0 n 16 n=0

3 neodređena: n=0 n= 16 n