ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\) \(\Rightarrow3x^2+4x+1=3a^2-b^2\)
Pt trở thành:
\(a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow a^2+2ab+b^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{5}}{2}b\\a=\frac{1-\sqrt{5}}{2}b< 0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x^2+2x}=\frac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\frac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow...\)
