<p>giúp e với ạ ( quan trọng)</p>

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
7,a,\ B=\frac{\sqrt{x}}{\sqrt{x} +2} +\frac{2\sqrt{x}}{\sqrt{x} -2} -\frac{3x+4}{x-4} \ \ \ \ ( x\geqslant 0;x\neq 4)\\
\ \ \ \ B=\frac{\sqrt{x} .\left(\sqrt{x} -2\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} +\frac{2\sqrt{x} .\left(\sqrt{x} +2\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} -\frac{3x+4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\
\ \ \ \ B=\frac{x-2\sqrt{x} +2x+4\sqrt{x} -3x-4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\
\ \ \ \ B=\frac{2\sqrt{x} -4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\
\ \ \ \ B=\frac{2.\left(\sqrt{x} -2\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\
\ \ \ \ B=\frac{2}{\sqrt{x} +2}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
b,\ \ \ \ B=\frac{2}{3}\\
\Leftrightarrow \frac{2}{\sqrt{x} +2} =\frac{2}{3}\\
\Leftrightarrow \sqrt{x} +2=3\\
\Leftrightarrow \sqrt{x} =1
\end{array}$
$\displaystyle \Leftrightarrow x=1$ (thỏa mãn điều kiện)