\(\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}\) ( ĐK: \(x\ge0\) )
\(\Rightarrow5\left(x+1\right)=\left(6-\sqrt{6}\right)\left(x+\sqrt{x}+1\right)\)
\(\Rightarrow5x+5=6x+6\sqrt{x}+6-x\sqrt{6}-\sqrt{6x}-\sqrt{6}\)
\(\Rightarrow x+1+6\sqrt{x}-x\sqrt{6}-\sqrt{6x}-\sqrt{6}=0\)
\(\Rightarrow\left(x+1\right)+\left(6\sqrt{x}-\sqrt{6x}\right)-\left(x\sqrt{6}+\sqrt{6}\right)=0\)
\(\Rightarrow\left(x+1\right)+\sqrt{6x}\left(\sqrt{6}-1\right)-\sqrt{6}\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(1-\sqrt{6}\right)-\sqrt{6x}\left(1-\sqrt{6}\right)=0\)
\(\Rightarrow\left(1-\sqrt{6}\right)\left(x+1-\sqrt{6x}\right)=0\)
Do đó: \(x-\sqrt{6x}+1=0\)
\(\Rightarrow x-2\sqrt{x}.\frac{\sqrt{3}}{\sqrt{2}}+\frac{3}{2}-\frac{3}{2}+1=0\)
\(\Rightarrow\left(\sqrt{x}-\frac{\sqrt{3}}{\sqrt{2}}\right)^2-\frac{1}{2}=0\)
\(\Rightarrow\left(\sqrt{x}-\frac{\sqrt{3}}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\left(x-\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{\sqrt{3}+1}{\sqrt{2}}\\\sqrt{x}=\frac{\sqrt{3}-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4+2\sqrt{3}}{2}\\x=\frac{4-2\sqrt{3}}{2}\end{matrix}\right.\) ( TMĐK)
