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 rồi tính

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

Rút gọn biểu thức

a) \(\sqrt {9{{(3 - a)}^2}} \) với \(a > 3\); b) \(\sqrt {{a^2}{{(a - 2)}^2}} \) với \(a < 0\).

Thực hiện phép tính

a) \(\sqrt {\frac{1}{8}}  \cdot \sqrt 2  \cdot \sqrt {125}  \cdot \sqrt {\frac{1}{5}} ;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt {\sqrt 2  - 1}  \cdot \sqrt {\sqrt 2  + 1} \).

b) \(\sqrt {{{(\sqrt 2  - 3)}^2}}  \cdot \sqrt {11 + 6\sqrt 2 } ;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt {{{(\sqrt 3  - 3)}^2}}  \cdot \sqrt {\frac{1}{{3 - \sqrt 3 }}} \).

\(c)\, - \frac{2}{3}\sqrt {\frac{{{{(a - b)}^2}{b^5}}}{c}}  \cdot \frac{9}{4}\sqrt {\frac{{{c^3}}}{{2(a - b)}}} \sqrt {98b} \)

d) \(\left( {\sqrt 6  - 3\sqrt 3  + 5\sqrt 2  - \frac{1}{2}\sqrt 1 } \right)2\sqrt 6 \)

e) \(\left( {\sqrt {ab}  + 2\sqrt {\frac{b}{a}}  - \sqrt {\frac{a}{b} + \sqrt {\frac{1}{{ab}}} } } \right)\sqrt {ab} \).

g) \(\left( {\frac{{am}}{b}\sqrt {\frac{n}{m}}  - \frac{{ab}}{n}\sqrt {mn}  + \frac{{{a^2}}}{{{b^2}}}\sqrt {\frac{m}{n}} } \right){a^2}{b^2} \cdot \sqrt {\frac{n}{m}} \).

Tìm \(x\) biết

a) \(\sqrt {9x} = 15\);       b) \(\sqrt {4{x^2}} = 8\)                                            c) \(\sqrt {4(x + 1)} = \sqrt 8 \);

d) \(\sqrt {9{{(2 - 3x)}^2}} = 6:\)   e) \(\sqrt {{x^2} - 4} - \sqrt {x - 2} = 0\).

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 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) \(\frac{{\sqrt 2  + \sqrt 3  + \sqrt 6  + \sqrt 8  + \sqrt {16} }}{{\sqrt 2  + \sqrt 3  + \sqrt 4 }}\)