Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

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

b,   x²-2xy+y²-z²

=(x-y)²-z²

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

con bai 32, 33 neu ban tra loi duoc minh h them

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

bn có phải Mod Yonnii bên hoidap247 ko ak?

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

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

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

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

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

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

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

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

d: \(x^2-xy+2y-2x\)

=x(x-y)-2(x-y)

=(x-y)(x-2)

c: \(x^2-6x-4y^2+9\)

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

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

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

b: \(x^3\left(2y-1\right)-125\left(2y-1\right)\)

\(=\left(2y-1\right)\left(x^3-125\right)\)

\(=\left(2y-1\right)\left(x-5\right)\left(x^2+5x+25\right)\)

a: \(5x^3y-10x^2y^2+5xy^3\)

\(=5xy\cdot x^2-5xy\cdot2xy+5xy\cdot y^2\)

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

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

1

\(x^3-64=x^3-4^3\)

\(\Rightarrow\left(x-4\right)\left(x^2+4x+4^2\right)\)

x³-64=x³-4³=(x-4)(x²+4x+16)

x^2 + 7x -15

= x^2 + 7x +12,25  -27,25

= (x+3,5)^2 - 27, 25

= ( x+3,5 - \(\sqrt{27,25}\))(x+3,5+\(\sqrt{27,25}\))

uk cảm ơn nha!!!!