2.4.tich phan ham_luong_giac_co_ban

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Text of 2.4.tich phan ham_luong_giac_co_ban

3

Bi 4. Tch phn c bn ca cc hm s lng gic

BI 4. TCH PHN C BN CA CC HM S LNG GICA. CNG THC S DNG1. KHAI TRIN NH THC NEWTON

trong v vi qui c 0! ( 12. CC CNG THC NGUYN HM LNG GIC

B. CC DNG TCH PHN

I. Dng 1:

1. Cng thc h bc

2. Phng php2.1. Nu n chn th s dng cng thc h bc2.2. Nu n ( 3 th s dng cng thc h bc hoc bin i theo 2.3.

2.3. Nu 3 ( n l (n ( 2p (1) th thc hin bin i:

II. Dng 2: (m, n(N)

1. Phng php:

1.1. Trng hp 1: m, n l cc s nguyn

a. Nu m chn, n chn th s dng cng thc h bc, bin i tch thnh tng.

b. Nu m chn, n l (n (2p (1) th bin i:

c. Nu m chn, n l (n (2p (1) th bin i:

d. Nu m l, n l th s dng bin i 1.2. hoc 1.3. cho s m l b hn.1.2. Nu m, n l cc s hu t th bin i v t u ( sinx ta c:

(*)

Tch phn (*) tnh c ( 1 trong 3 s l s nguyn 2. Cc bi tp mu minh ha

t ( ;

(

Cch 2:

III. Dng 3: (n(N)

1. Cng thc s dng

2. Cc bi tp mu minh ha

IV. Dng 4:

1. Phng php: Xt i din

1.1. Nu n chn (n ( 2k) th bin i:

1.2. Nu m l, n l (m ( 2k (1, n ( 2h (1) th bin i:

( y )

1.3. Nu m chn, n l (m ( 2k, n ( 2h ( 1) th s dng bin i:

H thc trn l h thc truy hi, kt hp vi bi tch phn hm phn thc hu t ta c th tnh c D4.1.

2. Cc bi tp mu minh ha:

EMBED Equation.DSMT4

3. Cc bi tp dnh cho bn c t gii:

V. Dng 5: S dng cng thc bin i tch thnh tng

1. Phng php:

2. Cc bi tp mu minh ha:

3. Cc bi tp dnh cho bn c t gii:

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