\(ĐKXĐ:\left[{}\begin{matrix}x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(\sqrt{\dfrac{2x+1}{x-1}}=a\ge0\Rightarrow\dfrac{2x+1}{x-1}=a^2\)
Phương trình trở thành :
\(a^2+a-3=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-3\right)=13>0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{13}}{2}\left(t.m\right)\\x=\dfrac{-1-\sqrt{13}}{2}\left(loai.vi.a\ge0\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{2x+1}{x-1}=\dfrac{7-\sqrt{13}}{2}\Leftrightarrow2+\dfrac{3}{x-1}=2+\dfrac{3-\sqrt{13}}{2}\\ \Leftrightarrow\dfrac{3}{x-1}=\dfrac{3-\sqrt{13}}{2}\Leftrightarrow x-1=\dfrac{6}{3-\sqrt{13}}\\ \Leftrightarrow x-1=\dfrac{-9-3\sqrt{13}}{2}\Leftrightarrow x=\dfrac{-7-3\sqrt{13}}{2}\)
