Bài 1:
a) \(=2xy\left(x+2-4x^2y\right)\)
b) \(=\left(x-2\right)\left(x+2\right)\)
c) \(=\left(x-y\right)^2-49=\left(x-y-7\right)\left(x-y+7\right)\)
d) \(=5\left(x+y\right)-z\left(x+y\right)=\left(x+y\right)\left(5-z\right)\)
Bài 2:
a) \(\Rightarrow x\left(x-\dfrac{1}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+4\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Bài 1:
\(a,=2x\left(x^2+2y-4x^2y\right)\\ b,=\left(x-2\right)\left(x+2\right)\\ c,=\left(x-y\right)^2-49=\left(x-y-7\right)\left(x-y+7\right)\\ d,=5\left(x+y\right)-z\left(x+y\right)=\left(5-z\right)\left(x+y\right)\)
Bài 2:
\(a,\Leftrightarrow x\left(x-\dfrac{1}{5}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
