Bài 2:
a: Ta có: \(5x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{2}{5}\end{matrix}\right.\)
b: Ta có: \(9\left(3x-2\right)=x\left(2-3x\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-9\end{matrix}\right.\)
Bài 1:
a: Ta có: \(x^2+2xy+y^2-16a^2\)
\(=\left(x+y\right)^2-\left(4a\right)^2\)
\(=\left(x+y+4a\right)\left(x+y-4a\right)\)
b: Ta có: \(x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-25\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
c: Ta có: \(x^2+3x-xy-3y\)
\(=x\left(x+3\right)-y\left(x+3\right)\)
\(=\left(x+3\right)\left(x-y\right)\)
d: Ta có: \(3a^2-4b+4a-3b^2\)
\(=3\left(a-b\right)\left(a+b\right)+4\left(a-b\right)\)
\(=\left(a-b\right)\left(3a+3b+4\right)\)
e: Ta có: \(x^3+x^2-x-1\)
\(=x^2\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-1\right)\)
\(=\left(x+1\right)^2\cdot\left(x-1\right)\)
f: Ta có: \(a^2+2ab+b^2-ac-bc\)
\(=\left(a+b\right)^2-c\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-c\right)\)
