c/ 3x-9xy-9y²+6y-1

=3x.(1-3y)-(1-6y+9y²)

=3x.(1-3y)-(1-3y)²

=(1-3y)[3x-(1-3y)]

=(1-3y)(3x-1+3y)

d/ x⁴+x²y²+y⁴

=x⁴+2x²y²+y⁴-x²y²

=(x²+y²)²-x²y²

=(x²-xy+y²)(x²+xy+y²)

\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)

\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

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

\(=4x\left(2x+1\right)\)

\(b,6x-6y-x^2+xy\)

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

\(=6\left(x-y\right)-x\left(x-y\right)\)

\(=\left(x-y\right)\left(6-x\right)\)

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 )

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

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

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

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

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

a) \(-y^2+\dfrac{1}{9}\)

\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)

\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)

b) \(4^4-256\)

\(=4^4-4^4\)

\(=0\)

c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)

\(=\left(3x-9\right)^2-\left(2x+2\right)^2\)

\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)

\(=\left(5x-7\right)\left(x-11\right)\)

\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)

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

(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 )

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

a,7(x+y)

b,2xy(x-3y)

chuc hk tot.

\(3x^4-48=3\left(x^4-16\right)=3\left[\left(x^2\right)^2-4^2\right]\\ =3\left(x^2-4\right)\left(x^2+4\right)\\ =3\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)