Giải phương trình: \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}\)
Help Hung nguyen;Ace Legona; Neet ; Mỹ Duyên;.................
\(Pt\Leftrightarrow x-\sqrt{\dfrac{x-1}{x}}=\sqrt{\dfrac{x^2-1}{x}}\)( DKXĐ : \(x\ge1\))
\(\Leftrightarrow x=\sqrt{\dfrac{x-1}{x}}\left(\sqrt{x+1}+1\right)=\sqrt{\dfrac{x-1}{x}}.\dfrac{x+1-1}{\sqrt{x+1}-1}\)
\(\Leftrightarrow x=\dfrac{\sqrt{x}.\sqrt{x-1}}{\sqrt{x+1}-1}\)
\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\\sqrt{x}=\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}\end{matrix}\right.\)
Có:\(x=\dfrac{x-1}{x+2-2\sqrt{x+1}}\) (ĐK: \(\sqrt{x+1}>1\))
\(\Leftrightarrow x^2+\left(x+1\right)-2x\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x-\sqrt{x+1}\right)^2=0\)
\(\Leftrightarrow x=\sqrt{x+1}\)
\(\Leftrightarrow x^2-x-1=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(chọn\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\)
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