p =x(x³ - 64) = x(x-4)( x² +4x +16)

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

(Áp dụng 7 hằng đẳng thức đáng nhớ)

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

2x²y - 4xy² + 6xy

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

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

\(P=x^4-64x=x\left(x^3-4^3\right)=x\left(x-4\right)\left(x^2+4x+16\right)\)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

a,7(x+y)

b,2xy(x-3y)

chuc hk tot.

Câu a:

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

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

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

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 )

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

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

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

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