TXĐ: \(x\le\dfrac{-7}{2};x\ge6;x=1\)
\(\sqrt{\left(x-1\right)\left(2x+7\right)}+\sqrt{\left(x-1\right)\left(3x-18\right)}=\sqrt{\left(x-1\right)\left(7x+1\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}\end{matrix}\right.\)
Pt1: \(\sqrt{x-1}=0\Rightarrow x=1\)
Pt2: \(\sqrt{2x+7}+\sqrt{3x-18}=\sqrt{7x+1}\)
\(\Leftrightarrow5x-11+2\sqrt{\left(2x+7\right)\left(3x-18\right)}=7x+1\)
\(\Leftrightarrow\sqrt{\left(2x+7\right)\left(3x-18\right)}=x+6\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\\left(2x+7\right)\left(3x-18\right)=\left(x+6\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\5x^2-27x-162=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{-18}{5}\end{matrix}\right.\)
Vậy pt có 3 nghiệm: \(\left[{}\begin{matrix}x=1\\x=9\\x=\dfrac{-18}{5}\end{matrix}\right.\)
