Tính:

a) \(\sqrt {1,2.270} ;\quad \sqrt {55.77.35} \).

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

c) \(\left( {\sqrt {\frac{8}{3}}  - \sqrt {24}  + \sqrt {\frac{{50}}{3}} } \right) \cdot \sqrt 6 ;\,\,\,\,\,\,\,\,\,\,\,{\left( {\sqrt {\frac{3}{2}}  - \sqrt {\frac{2}{3}} } \right)^2}\)

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

Rút gọn biểu thức

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

a) Cho \(a > 0.\) Chứng minh \(a + \frac{1}{a} \ge 2;\)

b) Cho \(a \ge 0,\;b \ge 0.\) Chứng minh \(\sqrt {\frac{{a + b}}{2}}  \ge \frac{{\sqrt a  + \sqrt b }}{2};\)

c) Cho \(a,b > 0.\) Chứng minh \(\sqrt a  + \sqrt b  \le \frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }};\)

d) Chứng minh \(\frac{{{x^2} + 2}}{{\sqrt {{x^2} + 1} }} \ge 2\) với mọi \(x.\)

Giải phương trình

a) \(\sqrt {\frac{{4x - 1}}{{x + 1}}}  = 3\);                                                                    b) \(\frac{{\sqrt {4x - 1} }}{{\sqrt {x + 1} }} = 3\);

c) \(\sqrt {49x - 98}  - 14\sqrt {\frac{{x - 2}}{{49}}}  = 3\sqrt {x - 2}  + 8\);                               d) \(\sqrt {25x - 25}  - \frac{{15}}{2}\sqrt {\frac{{x - 1}}{9}}  = 6 + \sqrt {x - 1} \).

Chứng minh đẳng thức

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

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) \(\sqrt {{x^2} - 4} - \sqrt {x - 2} = 0\).

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