Xin lỗi , mình làm nhầm . Làm lại nhé :
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) ( ĐKXĐ : x \(\ge\dfrac{1}{2}\) )
\(\Leftrightarrow x-\sqrt{2x-1}=2\)
\(\Leftrightarrow x-2=\sqrt{2x-1}\)
\(\Leftrightarrow x^2-4x+4=2x-1\)
\(\Leftrightarrow x^2-2x+1=4x-4\)
\(\Leftrightarrow\left(x-1\right)^2=4\left(x-1\right)\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=5\) ( tmđk )
Vậy phương trình có nghiệm x = 5 .
ĐK: \(x\ge\dfrac{1}{2}\)
Ta có
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|=2\) (*)
- TH1: \(\dfrac{1}{2}\le x< 1\) thì
(*) \(\Rightarrow1-\sqrt{2x-1}=2\) \(\Leftrightarrow\sqrt{2x+1}=-1\) (vô lý)
- TH2: \(x\ge1\) thì
(*) \(\Rightarrow\sqrt{2x-1}-1=2\) \(\Leftrightarrow\sqrt{2x-1}=3\) \(\Leftrightarrow x=5\) (Nhận)
Vậy x = 5
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) ( ĐKXĐ : \(x\ge\dfrac{1}{2}\) ; \(x\ne1\) )
\(\Leftrightarrow x-\sqrt{2x-1}=2\)
\(\Leftrightarrow x-2=\sqrt{2x-1}\)
\(\Leftrightarrow\left(x-2\right)^2=2x-1\)
\(\Leftrightarrow x^2-4x+4-2x+1=0\)
\(\Leftrightarrow x^2-5x-x+5=0\)
\(\Leftrightarrow x\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(ko-tm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 5 .