\(3x-7\sqrt{x}+4=3x-4\sqrt{x}-3\sqrt{x}+4=\sqrt{x}\left(3\sqrt{x}-4\right)-\left(3\sqrt{x}-4\right)=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)

\(3x-7\sqrt{x}+4=3x-3\sqrt{x}-4\sqrt{x}+4\)

\(=3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)\)

\(=\left(3\sqrt{x}-4\right)\cdot\left(\sqrt{x}-1\right)\)

\(x+\sqrt{x}-12=\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

\(2x-5\sqrt[]{x}+3=2x-2\sqrt[]{x}-3\sqrt[]{x}+3=2\sqrt[]{x}\left(\sqrt[]{x}-1\right)-3\sqrt[]{x}\left(\sqrt[]{x}-1\right)=\left(2\sqrt[]{x}-3\sqrt[]{x}\right)\left(\sqrt[]{x}-1\right)\)

=\(\sqrt{x}\left(\sqrt{x}-1\right)\)

\(x+\sqrt{x}-12=\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

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

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

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