ĐKXĐ: \(x\ge-\frac{1}{3}\)
Do \(\sqrt{3x+1}+\sqrt{x+2}>0;\forall x\ge-\frac{1}{3}\)
Nhân 2 vế của pt với \(\sqrt{3x+1}+\sqrt{x+2}\) và rút gọn ta được:
\(\left(2x-1\right)\left(\sqrt{3x^2+7x+2}+4\right)=2\left(2x-1\right)\left(\sqrt{3x+1}+\sqrt{x+2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\sqrt{3x^2+7x+2}+4=2\left(\sqrt{3x+1}+\sqrt{x+2}\right)\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow\sqrt{\left(3x+1\right)\left(x+2\right)}-2\sqrt{3x+1}-2\left(\sqrt{x+2}-2\right)=0\)
\(\Leftrightarrow\sqrt{3x+1}\left(\sqrt{x+2}-2\right)-2\left(\sqrt{x+2}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{3x+1}-2\right)\left(\sqrt{x+2}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=2\\\sqrt{x+2}=2\end{matrix}\right.\) \(\Leftrightarrow...\)
