\(3x^2y-6xy+2x-4\) (sửa đề)

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

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

\(=2x\left(x-3y\right)+5\left(x-3y\right)=\left(x-2y\right)\left(2x+5\right)\)

(x−2y)(2x+5)

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

=3y*(x^2-3yz+4x)

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 

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

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

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

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

= (x³ + 3x²y + 3xy² + y³) - (x+y)

= (x + y)³ - (x + y)

= (x + y).[(x+y)² - 1 ]

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

\(5x^2-6xy+y^2=\left(9x^2-6xy+y^2\right)-4x^2=\left(3x-y\right)^2-4x^2=\left(3x-y-2x\right)\left(3x-y+2x\right)=\left(x-y\right)\left(5x-y\right)\)

\(5x^2-6xy+y^2\)

\(=5x^2-5xy-xy+y^2\)

\(=5x\left(x-y\right)-y\left(x-y\right)\)

\(=\left(x-y\right)\left(5x-y\right)\)

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

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

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

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

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

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