Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

=(x-y)(x+y) - (x+y)

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

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

=4x²-y²+8y-16

=4x²-(y² - 8y+16)

=4x²-(y-4)²

=(2x+y-4)(2x-y+4)

   4x²-y²+8(y-2)

=4x²-y²+8.y-8.2

=4x²-y²+8y-16

4x(x+y)(x+y+z)(x+z)+(yz)^2

=(2x(x+y+z))(2(x+y)(x+z)+(yz)^2

=(2x^2+2xy+2xz)(2x^2+2xy+2xz+2yz)+(yz)^2

Đặt t=C

=(t-yz)(t+yz)-(yz)^2

=t^2-(yz)^2+(yz)^2=t^2=(2x^2+2xy+2xz+yz)^2

=

cam on ban nhung mk lam xong roi

\(4x^2-y^2+8\left(y-2\right)=4x^2-y^2+8y-16=4x^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)

\((4x-y)(a+b)(4x-y)(c-1)\)

\(=\left(4x-y\right)\left(4x-y\right)=\left(4x-y\right)^{1+1}=\left(4y-2\right)^2\)

\(=\left(a+b\right)\left(4x-y\right)^2\left(c-1\right)\)

(4x-y)(a+b)(4x-y)(c-1)

= ( 4x - y ) ( 4x - y ) = ( 4x - y ) ^{1 + 1} = ( 4y - 2 ) ²

= (a + b ) ( 4x - y )²  ( c - 1 )

Bài giải : 

(4x-y)(a+b)(4x-y)(c-1)

= ( 4x - y ) ( 4x - y ) = ( 4x - y ) ^{1 + 1} 

= ( 4y - 2 ) ²

= (a + b ) ( 4x - y )²  ( c - 1 )

Lưu ý rằng ba điều kiện đầu tiên yếu tố như (x + 1) ^ 2, do đó chúng ta có:
x^2 + 2x + 1 - y^2 = (x + 1)^2 - y^2. 

 (x + 1)^2 - y^2 = [(x + 1) + y][(x + 1) - y], từ a^2 - b^2 = (a + b)(a - b) 
= (x + y + 1)(x - y + 1).