đkxđ: `x≥0; x\ne 1`
`B=\sqrtx=(\sqrtx+1)/(\sqrtx-1)`
`⇔\sqrtx(\sqrtx-1)=\sqrtx+1`
`⇔x-\sqrtx=\sqrtx+1`
`⇔x-2\sqrtx-1=0`
`⇔(\sqrtx-1)^2-2=0`
`⇔(\sqrtx-\sqrt2-1)(\sqrtx+\sqrt2-1)=0`
\(⇔\left[ \begin{array}{l}\sqrt{x}-\sqrt{2}-1=0\\\sqrt{x}+\sqrt{2}-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}\sqrt{x}=1+\sqrt{2}\\\sqrt{x}=1-\sqrt{2}<0(loại)\end{array} \right.\)
`⇔\sqrtx=1+\sqrt2`
`⇔x=3+2\sqrt2`
Vậy `x=3+2\sqrt2`
