Question

giải pt \(\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}=x\)

Answer 1

ĐK: \(-1\le x< 0;x\ge1\)

TH1: \(-1\le x< 0\Rightarrow VP< 0;VT\ge0\Rightarrow\) vô nghiệm

TH2: \(x\ge1\)

\(pt\Leftrightarrow x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}\)

\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\)

\(\Leftrightarrow x^2-x+1-2\sqrt{x^2-x}=0\)

\(\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{x^2-x}=1\)

\(\Leftrightarrow x^2-x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)

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