\(\sqrt{5-3x}+\sqrt{x+1}=\sqrt{3x^2-4x+4}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}5-3x\ge0\\x+1\ge0\\3x^2-4x+4\ge0\end{matrix}\right.\)(*)
(1)\(\Leftrightarrow5-3x+x+1+2\sqrt{\left(5-3x\right)\left(x+1\right)}=3x^2-4x+4\)
\(\Leftrightarrow2\sqrt{\left(5-3x\right)\left(x+1\right)}=3x^2-2x-2\)
\(\Leftrightarrow2\sqrt{-\left(3x^2-2x-2\right)+3}=3x^2-2x-2\)
Đặt \(3x^2-2x-2=t\)
PTTT: \(2\sqrt{3-t}=t\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\le3\\4\left(3-t\right)=t^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}t\le3\\\left[{}\begin{matrix}t=2\left(tm\right)\\t=6\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(t=2\Leftrightarrow3x^2-2x-4=0\)(tự giải tiếp nha nhớ thay vào(*) xem nghiệm có thỏa mãn không với)
