Question

Phân tích các đa thức sau thành nhân tử : a) x^2-xy+3x-3y b) 2xy+3z+6y+xz c) x^2-6x-y^2+9 d)xy-z-xz+y

Answer 1

\(a,=x\left(x-y\right)+3\left(x-y\right)\\ =\left(x+3\right)\left(x-y\right)\\ b,=\left(2xy+6y\right)+\left(3z+xz\right)\\ =2y\left(x+3\right)+z\left(3+x\right)\\ =\left(2y+z\right)\left(x+3\right)\\ c,=\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\left(x-3+y\right)\\ d,=\left(xy+y\right)-\left(z+xz\right)\\ =y\left(x+1\right)-z\left(1+x\right)\\ =\left(y-z\right)\left(1+x\right)\)

Answer 2

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

b: =2x(y+z)+3(y+z)=(2x+3)(y+z)

c: =(x-3)²-y²

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

Answer 3

`a, x^2 - xy + 3x - 3y`

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

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

`b, 2xy + 3z + 6y + zx`

`= 2y(x + 3) + z(x+3)`

`= (2y+z)(x+3)`

`c, (x-3)^2 - y^2`

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

`d, x(y-z) + y-z`

`= (x+1)(y-z)`