\(\sqrt{2+\sqrt{3x-5}}=\sqrt{x+1}\left(ĐK:x\ge\frac{5}{3}\right)\)
\(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow3x-5=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-3x+5=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\left(TM\right)\\x=3\left(TM\right)\end{array}\right.\)
Vậy x=2; x=3
PT<=> \(\begin{cases}x+1\ge0\\\sqrt{3x+5}=x-1\end{cases}\)
<=>\(\begin{cases}x\ge1\\x\ge-1\\3x+5=x^2-2x+1\end{cases}\)
<=> \(\begin{cases}x\ge1\\x=\frac{5\pm\sqrt{41}}{2}\end{cases}\)
=> x=\(\frac{5+\sqrt{41}}{2}\)
