Lời giải:
ĐKXĐ:.........
Ta có:
\(\sqrt{1+\frac{1}{x+1}}+\sqrt{\frac{1}{x+1}}=\sqrt{x}+\frac{1}{\sqrt{x}}\)\(=\frac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow \sqrt{1+\frac{1}{x+1}}=\frac{x+1}{\sqrt{x}}-\sqrt{\frac{1}{x+1}}\)
Bình phương hai vế suy ra:
\(1+\frac{1}{x+1}=\frac{(x+1)^2}{x}+\frac{1}{x+1}-2\sqrt{\frac{x+1}{x}}\)
\(\Leftrightarrow 1=\frac{x^2+2x+1}{x}-2\sqrt{\frac{x+1}{x}}\)
\(\Leftrightarrow 0=\frac{x^2+x+1}{x}-2\sqrt{\frac{x+1}{x}}\)
\(\Leftrightarrow \frac{x^2+x+1}{x}=2\sqrt{\frac{x+1}{x}}\)
\(\Rightarrow x^2+x+1=2\sqrt{x(x+1)}\)
\(\Leftrightarrow (\sqrt{x^2+x}-1)^2=0\Rightarrow \sqrt{x^2+x}=1\)
\(\Rightarrow x^2+x=1\Leftrightarrow x^2+x-1=0\)
\(\Leftrightarrow x=\frac{-1\pm \sqrt{5}}{2}\)
Kết hợp với điều kiện xác định suy ra \(x=\frac{-1+\sqrt{5}}{2}\)
Vậy.........
\(\sqrt{1+\dfrac{1}{x+1}}+\sqrt{\dfrac{1}{x+1}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{\dfrac{2+x}{x+1}}-\dfrac{1}{\sqrt{x}}\right)+\left(\sqrt{\dfrac{1}{x+1}}-\sqrt{x}\right)=0\)
\(\Leftrightarrow\dfrac{x^2+x-1}{\left(x^2+x\right)\left(\sqrt{\dfrac{2+x}{x+1}}+\dfrac{1}{\sqrt{x}}\right)}-\dfrac{x^2+x-1}{\left(x+1\right)\left(\sqrt{\dfrac{1}{x+1}}+\sqrt{x}\right)}=0\)
