ĐK: \(x\ge\dfrac{1}{2}\)
\(pt\Leftrightarrow\sqrt{2x-1}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\2x-1=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2=-2\end{matrix}\right.\)
\(\Rightarrow\) Phương trình vô nghiệm
\(\sqrt{2x-1}-x-1=0\) (1)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
(1) \(\Rightarrow2x-1=\left(x+1\right)^2\)
\(\Leftrightarrow2x-1-x^2-2x-1=0\)
\(\Leftrightarrow-x^2-2=0\)
\(\Rightarrow\)PT vô nghiệm