Question

[Thắng Nguyễn]

Giải pt: x=\(\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\)

Answer 1

[Đặng Thanh Quang]

Đk \(x\ge1\)

Áp dụng bđt cosi có

\(\sqrt{x-\frac{1}{x}}=\sqrt{1\left(x-\frac{1}{x}\right)}\le\frac{1+x-\frac{1}{x}}{2}\)

\(\sqrt{1-\frac{1}{x}}=\sqrt{\frac{1}{x}\left(x-1\right)}\le\frac{\frac{1}{x}+x-1}{2}\)

\(\Rightarrow VT\le VP\)

Dấu = xay ra khi.........\(x=\frac{1+\sqrt{5}}{2}\)(do \(x\ge1\))

Answer 2

[Huy Hoang]

*ĐK* : \(\hept{\begin{cases}x\ne0\\x-\frac{1}{2}\ge0\\1-\frac{1}{x}\ge0\end{cases}\Leftrightarrow x\ge1}\)(1)

             \(x\ge0\)( điều kiện cần )

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

         \(\Leftrightarrow x\sqrt{x}=\sqrt{x-1}\left(\sqrt{x+1}+1\right)\)

          \(\Leftrightarrow x\sqrt{x}=\sqrt{x-1}.\frac{\left(x+1\right)-1}{\sqrt{x+1}-1}\)

          \(\Leftrightarrow\sqrt{x}.\left(\sqrt{x+1}-1\right)=\sqrt{x-1}\)( vì \(x\ge1>0\))

          \(\Leftrightarrow x\left(x+2-2\sqrt{x+1}\right)=x-1\)( vì \(x\ge1\)nên \(\sqrt{x+1}-1>0\))

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

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

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

          \(\Leftrightarrow x^2-x-x=0\Leftrightarrow x=\frac{1+\sqrt{5}}{2}\)hoặc \(x=\frac{1-\sqrt{5}}{2}\)

          \(\Leftrightarrow x=\frac{1+\sqrt{5}}{2}\)( vì đk \(x\ge1\))

Vậy nghiệm của PT trên là \(x=\frac{1+\sqrt{5}}{2}\)