\(=3^2-\left(x-y\right)^2=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]=\left(3-x+y\right)\left(3+x-y\right)\)

\(9-x^2+2xy-y^2\)

\(=9-\left(x^2-2xy+y^2\right)\)

\(=3^2-\left(x-y\right)^2\)

\(=\left(3-x+y\right)\left(3-x-y\right)\)

Giải:

a) \(3x^2y-6xy^2\)

\(=3xy\left(x-2y\right)\)

Vậy ...

b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)

\(=\left(2x-a\right)\left(x^2-y\right)\)

\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

Vậy ...

c) \(25a^2-c^2\)

\(=\left(5a-c\right)\left(5a+c\right)\)

Vậy ...

d) \(4-36x+81x^2\)

\(=2^2-2.2.9x+\left(9x\right)^2\)

\(=\left(2-9x\right)^2\)

Vậy ...

e) \(\left(x+7\right)2-\left(2x-9\right)2\)

\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)

\(=2\left(x+7-2x+9\right)\)

\(=2\left(16-x\right)\)

Vậy ...

f) \(x^2-6x+8\)

\(=x^2-6x+9-1\)

\(=\left(x-3\right)^2-1\)

\(=\left(x-4\right)\left(x-2\right)\)

Vậy ...

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

\(x^2-2xy+y^2-z^2\\=(x^2-2xy+y^2)-z^2\\=(x-y)^2-z^2\\=(x-y-z)(x-y+z)\)

Bài 1:

\(x^2-6x+9-y^2\)

\(=\left(x-3\right)^2-y^2\)

\(=\left(x-3+y\right)\left(x-3-y\right)\)

Bài 2:

\(x^2-x-12=0\)

\(\Leftrightarrow x^2-4x+3x-12=0\)

\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

1. x²+6x-9-y²

=-(x²-6x+y²)-3²

=-(x-y)²-3²

=(-x+y-3)(-x+y+3)

2. x² - x - 12 = 0

=> x² - 4x + 3x - 12 = 0

=> x( x - 4 ) + 3( x - 4 ) = 0

=> ( x + 3 )(x - 4 ) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+3=0\\x-4=0 \end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)

Vậy x=-3; x=4

dat y^2+y=z cho gon

\(z^2-9z+20=z^2-4z-5z+20=z\left(z-4\right)-5\left(z-4\right)=\left(z-4\right)\left(z-5\right)\)

\(thaylai:\left(y^2+y-4\right)\left(y^2+y-5\right)\)

thay hay thi "k" ko hay thi thoi 

x² - 3x + 3y - y²

= (x² - y²) - (3x - 3y)

= (x - y)(x + y) - 3(x - y)

= (x - y)(x + y - 3)

= x² - y² - 3x+3y = (x-y)(x+y) -3(x-y)

= (x+y+3)(x-y)

nhớ chọn cho mk nha!!!!!!