a,ĐK:\(x\ge\frac{3}{2}\)
\(PT\Leftrightarrow\left(3x+2\right)\sqrt{2x-3}-\left(3x+2\right)-2x^2+8=0\)
\(\Leftrightarrow\left(3x+2\right)\left(\sqrt{2x-3}-1\right)-2\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(3x+2\right).\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}-2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left[\frac{3x+2}{\sqrt{2x-3}+1}-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\frac{3x+2}{\sqrt{2x-3}+1}=x+2\left(1\right)\end{matrix}\right.\)
Giải (1)\(\Leftrightarrow3x+2=\sqrt{2x-3}\left(x+2\right)+x+2\)
\(\Leftrightarrow2x=\sqrt{2x-3}\left(x+2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\4x^2=\left(2x-3\right)\left(x^2+4x+4\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\2x^3+x^2-4x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left(x-2\right)\left(2x^2+5x+6\right)=0\end{matrix}\right.\) \(\Leftrightarrow x=2\left(tm\right)\)
Vậy \(x=2\)
b, Đề là \(5\sqrt{x+1}\) hay \(5\sqrt{x+4}\) vậy?
