Question

Giải PT: \(x=\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}\)

Answer 1

\(ĐK:x\ge1\\ PT\Leftrightarrow x-\sqrt{x-\dfrac{1}{x}}=\sqrt{1-\dfrac{1}{x}}\\ \Leftrightarrow x^2+x-\dfrac{1}{x}-2x\sqrt{x-\dfrac{1}{x}}=1-\dfrac{1}{x}\\ \Leftrightarrow x^2+x-1=2x\sqrt{x-\dfrac{1}{x}}\\ \Leftrightarrow x^4+x^2+1+2x^3-2x-2x^2=4x^3-4x\\ \Leftrightarrow x^4-2x^3-x^2+2x+1=0\\ \Leftrightarrow\left(x^2-x-1\right)^2=0\\ \Leftrightarrow x^2-x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=\dfrac{1+\sqrt{5}}{2}\)