\(ĐK:x\ge-\dfrac{5}{26}\)
Đặt \(a=\sqrt{26x+5};b=\sqrt{x^2+30}\left(a\ge0;b>0\right)\)
\(PT\Leftrightarrow\dfrac{a^2}{b}+2a=3b\\ \Leftrightarrow\left(a-b\right)\left(a+3b\right)=0\Leftrightarrow a=b\\ \Leftrightarrow\sqrt{26x+5}=\sqrt{x^2+30}\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=25\left(tm\right)\end{matrix}\right.\)
ĐKXĐ : \(x\ge-\dfrac{5}{26}\)
Đặt \(\sqrt{x^2+30}=a;\sqrt{26x+5}=b\)
Khi đó PT trở thành \(\dfrac{b^2}{a}+2b=3a\)
<=> \(b^2+2ab=3a^2\)
<=> \(\left(b-a\right)\left(b+3a\right)=0\)
<=> \(\left[{}\begin{matrix}b=a\\b=-3a\end{matrix}\right.\)
Khi b = a => \(\sqrt{x^2+30}=\sqrt{26x+5}\)
<=> x2 + 30 = 26x + 5
<=> (x - 1)(x - 25) = 0
<=> \(\left[{}\begin{matrix}x=1\\x=25\end{matrix}\right.\)(tm)
Khi b = -3a <=> \(\sqrt{26x+5}=-3\sqrt{x^2+30}\)
<=> 26x + 5 = -9(x2 + 30)
<=> 9x2 + 26x + 275 = 0
<=> \(9\left(x+\dfrac{13}{9}\right)^2+\dfrac{2306}{9}=0\)
<=> \(x\in\varnothing\)
Vậy x \(\in\left\{1;25\right\}\)
