ĐKXĐ: 2x-1>=0 và \(x-\sqrt{2x-1}>0\)
=>x>=1/2 và x>căn 2x-1
=>x>=1/2 và x^2>2x-1
=>x>=1/2 và x^2-2x+1>0
=>x>=1/2 và x<>1
\(\dfrac{1}{\sqrt[]{x-\sqrt[]{2x-1}}}\left(1\right)\)
\(\left(1\right)xđ\Leftrightarrow\left\{{}\begin{matrix}x-\sqrt[]{2x-1}>0\\2x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt[]{2x-1}< x\left(2\right)\\x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\left(I\right)\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x-1\ge0\\2x-1< x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge\dfrac{1}{2}\\x^2+2x-1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge\dfrac{1}{2}\\\left(x-1\right)^2>0,\forall x\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{1}{2}\)
\(\left(I\right)\Leftrightarrow x\ge\dfrac{1}{2}\)
Đính chính
\(...\left(x-1\right)^2>0,\forall x\ne1\)
\(...\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\end{matrix}\right.\)