Tìm \(x - y\) biết: \(\frac{1}{{\sqrt x }} + \frac{1}{{\sqrt y }} = 4 - \sqrt x - \sqrt y \)

Cho \(a \ge 0,b \ge 0,c \ge 0\). Chứng minh rằng:

a) \(\frac{{a + b}}{2} \ge \sqrt {ab} \) (bất đẳng thức Cauchy);

b) \(a + b + c \ge \sqrt {ab}  + \sqrt {bc}  + \sqrt {ca} \);

c) \(a + b + \frac{1}{2} \ge \sqrt a  + \sqrt b \).

Rút gọn biểu thức

a) \(\frac{{\sqrt {15}  - \sqrt 6 }}{{\sqrt {35}  - \sqrt {14} }}\);                                                    b) \(\frac{{\sqrt {10}  + \sqrt {15} }}{{\sqrt 8  + \sqrt {12} }}\)

c) \(\frac{{x + \sqrt {xy} }}{{y + \sqrt {xy} }}\) d) \(\frac{{\sqrt a  + a\sqrt b  - \sqrt b  - b\sqrt a }}{{ab - 1}}\)

e) \(\frac{{2\sqrt {15}  - 2\sqrt {10}  + \sqrt 6  - 3}}{{2\sqrt 5  - 2\sqrt {10}  - \sqrt 3  + \sqrt 6 }}\)                                                                    f) \(\frac{{\sqrt 2  + \sqrt 3  + \sqrt 6  + \sqrt 8  + \sqrt {16} }}{{\sqrt 2  + \sqrt 3  + \sqrt 4 }}\)

Chứng minh đẳng thức

a) \(\sqrt {9 - \sqrt {17} }  \cdot \sqrt {9 + \sqrt {17} }  = 8\);                                                      b) \(\left( {\frac{1}{{5 - 2\sqrt 6 }} + \frac{2}{{5 + 2\sqrt 6 }}} \right)\left( {15 + 2\sqrt 6 } \right) = 201\)

Tính: \(\left( {\frac{1}{2}\sqrt {\frac{1}{2}} - \frac{3}{2}\sqrt {4,5} + \frac{2}{5}\sqrt {50} } \right):\frac{4}{{15}}\sqrt {\frac{1}{8}} \).

Rút gọn biểu thức

a) \(\,\frac{{x\sqrt x  + y\sqrt y }}{{\sqrt x  + \sqrt y }} - {\left( {\sqrt x  - \sqrt y } \right)^2};\)

b) \(\,\,\sqrt {\frac{{x - 2\sqrt x  + 1}}{{x + 2\sqrt x  + 1}}} ,\,\,\left( {x \ge 0} \right)\)

c)\(\,\frac{{x - 1}}{{\sqrt y  - 1}}.\sqrt {\frac{{{{\left( {y - 2\sqrt y  + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^4}}}} ,\,\,\left( {x \ne 1,y \ne 1,y > 0} \right)\).

Rút gọn rồi tính

a) \(\sqrt {{{21,8}^2} - {{18,2}^2}} ;\)             b) \(\sqrt {{{6,8}^2} - {{3,2}^2}} ;\)                     c) \(\sqrt {{{146,5}^2} - {{109.5}^2} + 27 \cdot 256} \)