ĐKXĐ: \(x\ge0\)
Đặt \(\sqrt{x}+\sqrt{x+5}=a>0\)
\(\Rightarrow a^2=2x+5+2\sqrt{x\left(x+5\right)}\)
Phương trình trở thành:
\(a^2-5+a=25\)
\(\Leftrightarrow a^2+a-30=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-6\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\sqrt{x+5}=5\)
\(\Leftrightarrow\sqrt{x}-2+\sqrt{x+5}-3=0\)
\(\Leftrightarrow\frac{x-4}{\sqrt{x}+2}+\frac{x-4}{\sqrt{x+5}+3}=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x+5}+3}\right)=0\)
\(\Rightarrow x=4\)