Cho biểu thức \(P = \left( {\frac{1}{{\sqrt x + 1}} - \frac{{x + 2}}{{x\sqrt x + 1}}} \right):\frac{2}{{\sqrt x }}\).
a) Rút gọn \(P\).

b) Chứng minh rằng biểu thức \({\rm{P}}\) luôn luôn âm với mọi giá trị của \({\rm{x}}\) làm \({\rm{P}}\) xác định.

\(\begin{array}{l}a)\;\frac{1}{{34}}\sqrt {\frac{{{{165}^2} - {{124}^2}}}{{164}}} + 4\sqrt {\frac{{32}}{{{{176}^2} - {{112}^2}}}} = \frac{1}{{34}}\sqrt {\frac{{41.289}}{{164}}} + 4\sqrt {\frac{{32}}{{64.288}}} = \frac{1}{{34}}\sqrt {\frac{{289}}{4}} + 4\sqrt {\frac{1}{{576}}} \\ = \frac{1}{{34}} \cdot \frac{{17}}{2} + 4 \cdot \frac{1}{{24}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{{12}}\end{array}\)

b) \(\frac{{5\left( {\sqrt 6 - 1} \right)}}{{\sqrt 6 + 1}} + \frac{{\sqrt 2 - \sqrt 3 }}{{\sqrt 2 + \sqrt 3 }} = \frac{{5{{(\sqrt 6 - 1)}^2}}}{{6 - 1}} + \frac{{{{(\sqrt 2 - \sqrt 3 )}^2}}}{{2 - 3}} = 7 - 2\sqrt 6 - 5 + 2\sqrt 6 = 2.\)

a) Ta có \(\sqrt {\frac{9}{{16}}:\frac{{25}}{{36}}} - \sqrt {\frac{{49}}{8}} :\sqrt {3\frac{1}{8}} = \sqrt {\frac{9}{{16}}:\frac{{25}}{{36}}} - \sqrt {\frac{{49}}{8}:\frac{{25}}{8}} = \frac{3}{4}:\frac{5}{6} - \sqrt {\frac{{49}}{{25}}} = \frac{9}{{10}} - \frac{7}{5} = - \frac{1}{2}.\)

b) \(\sqrt {{{45,8}^2} - {{44,2}^2}} - \sqrt {6\left[ {{{(\sqrt 2 + 1)}^2} + {{(\sqrt 2 - 1)}^2}} \right]} = \sqrt {1,6.90} - \sqrt {6.6} = 4.3 - 6 = 6.\)