\(x^2-6xy+9y^2=x^2-2.\left[x.\left(3y\right)\right]+\left(3y\right)^2\)

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

1)P= x² - 6xy+9y²

= (x-3y)^2

2) P= x⁴ - 64x

= (x^3 - 64) x

= (x^3 - 4^3)x

= (x^3 - 12x^2 + 48x - 64)x

Phân tích theo cách nhóm hạng tử nhé bạn ! 

a, xy + xz - 2y - 2z

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

= (x - 2)(y + x)

b, x² - 6xy + 9y² - 25z²

= (x - 3y)² - 25z²

= (x - 3y - 5z)(x - 3y + 5z) 

Ngoài ra bạn có thể hỏi mình để bổ sung kiến thức nâng cao !

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 

Câu a:

\(x^2-y^2-x+3y-2=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2.y.\frac{3}{2}+\frac{9}{4}\right)\)

\(< =>\left(x-\frac{1}{2}\right)^2-\left(y-\frac{3}{2}\right)^2\)

\(< =>\left(x-\frac{1}{2}+y-\frac{3}{2}\right)\left(x-\frac{1}{2}-y+\frac{3}{2}\right)=\left(x+y-2\right)\left(x-y+1\right)\)

\(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)

2x²y - 4xy² + 6xy

= 2xy . ( x - 2y + 3 )

\(2x^2y-4xy^2+6xy=2xy\cdot\left(x-2y+3\right)\)

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

b) 9 - ( x-  y)^2 = ( 3 )^2 - ( x- y)^2

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

a/ \(3x^2y-6xy^2\)\(=3xy\left(x-2y\right)\) ( đây là p²   đặt nhân tử chung )

b/9-(x -y )²    =( 3 -x +y ) ( 3 + x+y )   ( dùng hđt số 3 để giải )

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

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

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

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

\(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-16\right)\)

\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

a,7(x+y)

b,2xy(x-3y)

chuc hk tot.