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Với \(a,b\in\mathbb{Z};a,b\ne0;a\ne3b;a\ne-5b\), ta có:
\(E=\dfrac{b\left(2a^2+10ab+a+5b\right)}{a-3b}:\dfrac{a^2b+5ab^2}{a^2-3ab}\)
\(=\dfrac{b\left[2a\left(a+5b\right)+\left(a+5b\right)\right]}{a-3b}:\dfrac{ab\left(a+5b\right)}{a\left(a-3b\right)}\)
\(=\dfrac{b\left(2a+1\right)\left(a+5b\right)}{a-3b}:\dfrac{b\left(a+5b\right)}{a-3b}\)
\(=\dfrac{b\left(2a+1\right)\left(a+5b\right)}{a-3b}\cdot\dfrac{a-3b}{b\left(a+5b\right)}\)
\(=2a+1\)
Vì \(2a+1\) là số nguyên lẻ với mọi a nguyên
nên \(E\) là số nguyên lẻ.
\(\text{#}Toru\)
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
Phân tích đa thức thành nhân tử
a. 3ab ( x+ y) - 6ab ( y+ x)
=( x + y) ( 3ab - 6ab )
= ( x +y ) ( - 3ab)
b.7a (x - 3)+a2(x2 - 9)
=7a( x- 3) + a2 ( x2 - 32)
=7a ( x - 3 ) + a2 ( x- 3 ) ( x+3 )
= ( x- 3) . 7a + a2 ( x + 3)
= ( x- 3) ( 7a +a2x + 3a2)
c. 34 (x + y) -x -y
= 34 ( x+ y) - ( x+y)
=(x +y ) ( 34 - 1) = 33 ( x+ y)
d. 25 x4 - 942
=( 5x2 )2 - 942
=( 5x2 - 94 ) ( 5x2+94)
e.( 5a - b )2 - ( 2a +3b)2
=( 5a -b -2a - 3b) (5a -b + 2a + 3b)
=(3a - 4b) (7a+ 2b)
k. 22 -3a - b2 +3b
=( 22 - b2 ) + ( -3a +3b)
=( 2-b) (2+b) + 3( -a +b)
a. (a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2
b. (a+b)^3= (a+b)(a+b)(a+b) = (a^2 + 2ab + b^2)(a + b) = a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 = a^3 + 3a^2b + 3b^2a + b^3
c. (a-b)^3= (a - b)(a-b)(a-b) = (a^2 - 2ab + b^2)(a - b) = a^3 - a^2b - 2a^2b + 2ab^2 + b^2a - b^3 = a^3 - 3a^2b + 3ab^2 - b^3
e. (a-b) ( a^2 + ab +b^2) = a^3 + a^2b + b^2a - ba^2 - ab^2 - b^3 = a^3 - b^3
g. ( a-b) ( a+b) = a^2 +ab -ab - b^2 = a^2 - b^2
a) \(\left(A+B\right)^2=\left(A+B\right)\left(A+B\right)=A^2+AB+AB+B^2=A^2+2AB+B^2\)
b) \(\left(A+B\right)^3=\left(A+B\right)^2\left(A+B\right)=\left(A^2+2AB+B^2\right)\left(A+B\right)\)( NHÂN ra nốt hộ mk nha ) :D !
c)\(\left(A+B\right)\left(A-B\right)=A^2+AB-AB-B^2=A^2-B^2\)
ý d tương tự nha :D !
Ta có : \(a+b=2\)
\(\Rightarrow\)\(a = 2 -b\)
\(A = 2a^2 +3b^2 +3ab\)
\(A = 2a^2 + 3b. (a+b)\)
\(A = 2. (2-b)^2+3b. (2-b+b)\)
\(A = 2. ( b^2 -4b+4)+6b\)
\(A = 2b^2 -8b+8+6b\)
\(A = 2b^2 -2b+8\)
\(A = 2. ( b ^2 -b+4)\)
\(A=2. (b^2 -2.b.{1\over2}+({1\over2})^2-({1\over2})^2+4)\)
\(A = 2. [ (b -{1\over2})^2-{15\over4}]\)
\(A =2. (b-{1\over2})^2 + {15\over2}\)\(\ge\)\({15\over2}\)
\(Min A ={15\over2}\)\(\Leftrightarrow\)\(a = {3\over2};b={1\over2}\)
Ta có : a+b=2→b=2−a
→P=2a2+3b2+3ab=2a2+3b(a+b)=2a2+3b.2=2a2+6b=2a2+6(2−a)=2a2−6a+12
→P=2(a2−3a)+12
→P=2(a2−2a.32+94)+152
→P=2(a−32)2+152≥152
→GTNNP=152
Dấu = xảy ra khi a−32=0
1) \(\left(a+b\right)^2\)
\(=\left(a+b\right)\left(a+b\right)\)
\(=a^2+ab+ab+b^2\)
\(=a^2+2ab+b^2\left(dpcm\right)\)
2) \(\left(a-b\right)^3\)
\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)\)
\(=\left(a^2-ab-ab+b^2\right)\left(a-b\right)\)
\(=\left(a^2-2ab+b^2\right)\left(a-b\right)\)
\(=a^3-a^2b-2a^2+2ab^2+ab^2-b^3\)
\(=a^3-3a^2b+3ab^2-b^3\left(dpcm\right)\)