ĐKXĐ: \(x\ge-1\)
\(\left(2x+3\right)\sqrt{2x+2}=\left(2x+3\right)\left(4x^2-7x-8\right)\)
\(\Leftrightarrow\sqrt{2x+2}=4x^2-7x-8\)
\(\Leftrightarrow\sqrt{2x+2}=4x^2+x-4\left(2x+2\right)\)
Đặt \(\sqrt{2x+2}=y\ge0\)
\(\Rightarrow y=4x^2+x-4y^2\)
\(\Leftrightarrow4\left(x^2-y^2\right)+x-y=0\)
\(\Leftrightarrow\left(x-y\right)\left(4x+4y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\4y=-4x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+2}=x\left(x\ge0\right)\\4\sqrt{2x+2}=-4x-1\left(x\le-\dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=x^2\left(x\ge0\right)\\16\left(2x+2\right)=16x^2+8x+1\left(x\le-\dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}+1\\x=\dfrac{3-2\sqrt{10}}{4}\end{matrix}\right.\)
