Điều kiện xác định $x\ge 0$
$\begin{array}{l} A = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt {x + 1} .\sqrt {x + 1} }} = \dfrac{{x - 1}}{{x + 1}}\\ B = \dfrac{1}{{\sqrt x + 1}} + \dfrac{{2\sqrt x }}{{x\sqrt x + \sqrt x + x + 1}}\\ B = \dfrac{1}{{\sqrt x + 1}} + \dfrac{{2\sqrt x }}{{x\left( {\sqrt x + 1} \right) + \left( {\sqrt x + 1} \right)}}\\ B = \dfrac{1}{{\sqrt x + 1}} + \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}\\ B = \dfrac{{x + 1 + 2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}} = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}\\ B = \dfrac{{\sqrt x + 1}}{{x + 1}}\\ b)A = 7B\\ \Leftrightarrow \dfrac{{7\sqrt x + 7}}{{x + 1}} = \dfrac{{x - 1}}{{x + 1}}\\ \Leftrightarrow x - 1 = 7\sqrt x + 7\\ \Leftrightarrow x - 7\sqrt x - 8 = 0\\ \Leftrightarrow \left( {\sqrt x + 1} \right)\left( {\sqrt x - 8} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt x + 1 = 0(L)\\ \sqrt x = 8 \Leftrightarrow x = 64 \end{array} \right.\\ \Rightarrow x = 64 \end{array}$
