E = ( 2x - 1 )( 4x² + 2x + 1 ) - 2x( 2x - 7 )( 2x + 7 )

= ( 2x )³ - 1³ - 2x( 4x² - 49 )

= 8x³ - 1 - 8x³ + 98x = 98x - 1

a, \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)

b, \(x^2-4x+4=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)

c, \(x^2-2x-3=\left(x-1\right)^2-4=\left(x-3\right)\left(x+1\right)\)

c, Đặt \(x^2-3x-1=t\)

\(t^2-12t+27=t^2-3t-9t+27=\left(t-9\right)\left(t-3\right)\)

Theo cách đặt \(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)

a) ĐK : a khác 0 , a khác b

\(=\frac{\left(a+b\right)\left(a-b\right)}{a\left(a-b\right)}-\frac{a^2}{a\left(a-b\right)}-\frac{b^2}{a\left(a-b\right)}\)

\(=\frac{a^2-b^2-a^2-b^2}{a\left(a-b\right)}=-\frac{2b^2}{a\left(a-b\right)}\)

b) ĐK : x khác +-y

\(=\frac{x\left(x^2-y^2\right)}{\left(x-y\right)\left(x+y\right)}-\frac{xy\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}-\frac{x^3}{\left(x-y\right)\left(x+y\right)}\)

\(=\frac{x^3-xy^2-x^2y+xy^2-x^3}{\left(x-y\right)\left(x+y\right)}=-\frac{x^2y}{\left(x-y\right)\left(x+y\right)}\)

c) ĐK : x khác 2

\(=\frac{2x+8}{\left(x-2\right)^2}-\frac{7\left(x-2\right)}{\left(x-2\right)^2}=\frac{2x+8-7x+14}{\left(x-2\right)^2}=\frac{22-5x}{\left(x-2\right)^2}\)

d) ĐK : a khác 1

\(=\frac{3a^2+3}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{3a^2+3-a^2+2a-1+2a^2+2a+2}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{4a^2+4a+4}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{4\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{4}{a-1}\)

Áp dụng hằng đẳng thức: (A-B)^2 :))

A: 3x+1

B: 3x+5

C = ( 3x + 1 )² - 2( 3x + 1 )( 3x + 5 ) + ( 3x + 5 )²

= [ ( 3x + 1 ) - ( 3x + 5 ) ]²

= ( 3x + 1 - 3x - 5 )²

= (-4)² = 16

4m²+m=5n²+n

{=}5m²+m=5n²+n+m²

{=}5(m²-n²)+(m-n)=m²

{=}(m-n)(5m+5n+1)=m²

là sao

đó là toán 8 đó hả 

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

<=> [ ( x² - 2 ) - 2( x - 1 ) ]² = 0

<=> ( x² - 2 - 2x + 2 )² = 0

<=> [ x(x-2) ]² = 0

<=> x = 0 hoặc x = 2