Cách 1 : \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Cách 2 : \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x^2+4x=8x+8\)
\(\Leftrightarrow4x^2+4x-8x-8=0\)
\(\Leftrightarrow4x^2-4x-8=0\)
\(\Leftrightarrow4x^2-4x+1-9=0\)
\(\Leftrightarrow\left(2x-1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=-3\\2x-1=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
4x(x+1)-8(x+1)=0
⇔ (x+1)(4x-8)=0
⇔4(x+1)(x-2)=0
⇔\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
vậy x=-1hoặc x=2
