(1,0 điểm) Rút gọn biểu thức $\mathrm{A}=\left(\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}+2}-\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}-2}+\frac{4\sqrt{\mathrm{x}}-1}{\mathrm{x}-4}\right):\frac{1}{\sqrt{\mathrm{x}}+2}$  với $\mathrm{x}\ge 0,\mathrm{x}\ne 4.$

Với \(x \ge 0,\,\,x \ne 4,\) ta có:

\(A = \left( {\frac{{\sqrt x }}{{\sqrt x + 2}} - \frac{{\sqrt x }}{{\sqrt x - 2}} + \frac{{4\sqrt x - 1}}{{x - 4}}} \right):\frac{1}{{\sqrt x + 2}}\)

 \( = \left[ {\frac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} - \frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} + \frac{{4\sqrt x - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}} \right]:\frac{1}{{\sqrt x + 2}}\)

 \[ = \frac{{\sqrt x \left( {\sqrt x - 2} \right) - \sqrt x \left( {\sqrt x + 2} \right) + 4\sqrt x - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\frac{1}{{\sqrt x + 2}}\]

 \[ = \frac{{x - 2\sqrt x - x - 2\sqrt x + 4\sqrt x - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\frac{1}{{\sqrt x + 2}}\]

 \[ = \frac{{ - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} \cdot \frac{{\sqrt x + 2}}{1} = - \frac{1}{{\sqrt x - 2}}.\]

Vậy với \(x \ge 0,\,\,x \ne 4\) thì \[A = - \frac{1}{{\sqrt x - 2}}.\]