ĐKXĐ: \(\left\{{}\begin{matrix}2x+5\ge0\\3x-5\ge0\end{matrix}\right.\Leftrightarrow x\ge\dfrac{5}{3}\)
Pt \(\Leftrightarrow\sqrt{2x+5}=\sqrt{3x-5}+2\)
\(\Leftrightarrow2x+5=3x-1+4\sqrt{3x-5}\)
\(\Leftrightarrow6-x=4\sqrt{3x-5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}6-x\ge0\\\left(6-x\right)^2=16\left(3x-5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le6\\x^2-12x+36=48x-80\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le6\\x^2-60x+116=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le6\\\left[{}\begin{matrix}x=2\\x=58\end{matrix}\right.\end{matrix}\right.\)
<=> x = 2 (tm)
Vậy pt có tập nghiệm là: S = {2}.
