a) \(2x^3+6xy-x^2z-3yz\)

= \(\left(2x^3+6xy\right)-\left(x^2z+3yz\right)\)

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

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

b)\(x^2-6xy+9y^2-49\)

=\(x^2-2.x.3y+\left(3y\right)^2-7^2\)

=\(\left(x-3y\right)^2-7^2\)

=\(\left(x-3y+7\right)\left(x-3y-7\right)\)

Cứu với mn ơi , mai em có ktra ạ 

x²-6xy+9y²-36 =(x²-6xy+9y²)-36 =(x-3y)²-6² =(x-3y+6)(x-3y-6)

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

b. \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

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

b: \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

a) Ta có: \(x^2y^2-x^2+6xy-9y^2\)

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

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

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

b) Ta có: \(9-x^2+2xy-y^2\)

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

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

\(=\left(9-x+y\right)\left(9+x-y\right)\)

x^2+6xy+9y^2-3x-9y+2

=( x^2+6xy+9y^2)-3(x+3y)+9/4 -1/4

=(x+3y)^2-3(x+3y)+(3/2)^2- 1/4

=(x+3y+3/2)^2-(1/2)^2

=(x+3y+3/2+1/2)(x+3y+3/2-1/2)=(x+3y+2)(x+3y+1)

\(3x^2-6xy+9y^2-12\)

\(=3\cdot x^2-3\cdot2xy+3\cdot3y^2-3\cdot4\)

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

=3*x^2-3*2xy+3*3y^2-3*4

=3(x^2-2xy+3y^2-4)

=x^2 + 2 * x * 3y + (3y)^2

=(x + 3y)^2

a) 2x³ + 6xy - x²z - 3yz

= ( 2x³ + 6xy ) - ( x²z + 3yz )

= 2x( x² + 3y ) - z( x² + 3y )

= ( x² + 3y )( 2x - z )

b) x² - 6xy + 9y² - 49

= ( x² - 6xy + 9y² ) - 49

= ( x - 3y )² - 7²

= ( x - 3y - 7 )( x - 3y + 7 )

c) x³ + 4x² + 16x + 64

= ( x³ + 4x² ) + ( 16x + 64 )

= x²( x + 4 ) + 16( x + 4 ) 

= ( x + 4 )( x² + 16 )

a) =(2x^3-x^2z)+(6xy-3yz)

=x^2(2x-z)+3y(2x-z)

=(x^2+3y)(2x-z)

b) =(x^2-6xy+9y^2)-7^2

=(x-3y)^2-7^2

=(x-3y+7)(x-3y-7)

c) =(x^3+4x^2)+(16x+64)

=x^2(x+4)+16(x+4)

=(x^2+16)(x+4)

\(x^4+6x^3+13x^2+12x+4\)

\(=x^4+x^3+5x^3+5x^2+8x^2+8x+4x+4\)

\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+8x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+5x^2+8x+4\right)\)

\(=\left(x+1\right)\left(x^3+x^2+4x^2+4x+4x+4\right)\)

\(=\left(x+1\right)\left[x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\right]\)

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