ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(x^2-2x-2\sqrt{2x-1}=0\)
\(\Leftrightarrow x^2-\left(2x-1+2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow x^2-\left(\sqrt{2x-1}+1\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2x-1}-1\right)\left(x+\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow x-\sqrt{2x-1}-1=0\) (do \(x+\sqrt{2x-1}+1>0;\forall x\ge\dfrac{1}{2}\))
\(\Leftrightarrow\sqrt{2x-1}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\2x-1=\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2-\sqrt{2}< 1\left(loại\right)\\x=2+\sqrt{2}\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=2+\sqrt{2}\)
TK:
\[
x^2 - 2x - 2\sqrt{2}x + 1 = 0
\]
\[
x^2 - (2 + 2\sqrt{2})x + 1 = 0
\]
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
\[
D = b^2 - 4ac
\]
\[
D = (-(2 + 2\sqrt{2}))^2 - 4 \cdot 1 \cdot 1
\]
\[
D = (2 + 2\sqrt{2})^2 - 4
\]
\[
D = (4 + 8\sqrt{2} + 8) - 4 = 8 + 8\sqrt{2}
\]
\[
x = \frac{2 + 2\sqrt{2} \pm \sqrt{8 + 8\sqrt{2}}}{2}
\]
\[
x = 1 + \sqrt{2} \pm \frac{\sqrt{8(1 + \sqrt{2})}}{2}
\]
\[
x_1 = 1 + \sqrt{2} + \sqrt{2(1 + \sqrt{2})}
\]
\[
x_2 = 1 + \sqrt{2} - \sqrt{2(1 + \sqrt{2})}
\]
\(x^2-2x=2\sqrt{2x-1}\left(x\le0;x\ge2\right)\)
\(\Rightarrow x^4-4x^3+4x^2=8x-4\)
\(\Rightarrow x^4-4x^3+4x^2-8x+4=0\)
\(\Rightarrow\left(x^2+2\right)\left(x^2-4x+2\right)=0\)
\(\Rightarrow x^2-4x+2=0\) (vì \(x^2+2>0\))
\(\Rightarrow x^2-4x+4=2\)
\(\Rightarrow\left(x-2\right)^2=2\)
\(\Rightarrow x-2=\pm\sqrt{2}\)
\(\Rightarrow x=2\pm\sqrt{2}\)
