Với \(x \ge 0,\,\,x \ne 1,\) ta có:

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

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

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

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

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

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

Vậy với \(x \ge 0,\,\,x \ne 1\) thì \(B = \frac{{\sqrt x }}{{\sqrt x + 1}}.\)

Trong tam giác M N P vuông tại N , sin góc N P M bằng