ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{26x+5}=a\ge0\\\sqrt{x^2+30}=b>0\end{matrix}\right.\)
\(\Rightarrow\frac{a^2}{b}+2a=3b\)
\(\Leftrightarrow a^2+2ab-3b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+3b\right)=0\)
\(\Leftrightarrow a-b=0\)
\(\Leftrightarrow\sqrt{26x+5}=\sqrt{x^2+30}\)
\(\Leftrightarrow x^2-26x+25=0\Rightarrow\left[{}\begin{matrix}x=1\\x=25\end{matrix}\right.\)