Bài 2:
a) Ta có: \(\dfrac{2}{3}x^2y-2xy^2+4xy\)
\(=2xy\cdot\dfrac{1}{3}x-2xy\cdot y+2xy\cdot2\)
\(=2xy\left(\dfrac{1}{3}x-y+2\right)\)
b) Ta có: \(4x^2-2x-3y-9y^2\)
\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-1\right)\)
Bài 3:
a) Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b) Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=2-5x\)
\(\Leftrightarrow x^3+27-x^3-2x-2+5x=0\)
\(\Leftrightarrow3x=-25\)
hay \(x=-\dfrac{25}{3}\)
c) Ta có: \(x^2-25=6x-9\)
\(\Leftrightarrow x^2-25-6x+9=0\)
\(\Leftrightarrow x^2-6x-16=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
