ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\) phương trình trở thành:
\(a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow a^2+2ab+b^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{5}}{2}b\\a=\frac{1-\sqrt{5}}{2}b< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow2\sqrt{x^2+2x}=\left(1+\sqrt{5}\right)\sqrt{2x-1}\)
\(\Leftrightarrow4x^2+8x=\left(12+4\sqrt{5}\right)x+6+2\sqrt{5}\)
\(\Leftrightarrow2x^2-2\left(1+\sqrt{5}\right)x+3+\sqrt{5}=0\)
\(\Delta'=\left(1+\sqrt{5}\right)^2-2\left(3+\sqrt{5}\right)=0\)
\(\Rightarrow x=\frac{1+\sqrt{5}}{2}\)
