Thực hiện phép tính.

a) \(\sqrt {1\frac{9}{{16}}} \);

b) \(\sqrt {\frac{{25}}{{64}}} \);

c) \(\sqrt {\frac{{230}}{{2,3}}} \).

Đáp án đúng là: A

\(C = \left( {\frac{{1 - \sqrt 2 }}{{1 + \sqrt 2 }} - \frac{{1 + \sqrt 2 }}{{1 - \sqrt 2 }}} \right):\sqrt {72} = \left( {\frac{{1 - \sqrt 2 }}{{1 + \sqrt 2 }} - \frac{{1 + \sqrt 2 }}{{1 - \sqrt 2 }}} \right).\frac{1}{{\sqrt {72} }}\)

\( = \left[ {\frac{{{{\left( {1 - \sqrt 2 } \right)}^2} - {{\left( {1 + \sqrt 2 } \right)}^2}}}{{\left( {1 + \sqrt 2 } \right)\left( {1 - \sqrt 2 } \right)}}} \right].\frac{1}{{6\sqrt 2 }} = \frac{{1 - 2\sqrt 2 + 2 - 1 - 2\sqrt 2 - 2}}{{ - 1}}.\frac{1}{{6\sqrt 2 }}\)

$=\frac{4\sqrt{2}}{6\sqrt{2}}=\frac{2}{3}$

Đáp án đúng là: A

Ta có: \(A = \sqrt {\frac{{\sqrt a - 1}}{{\sqrt b + 1}}} :\sqrt {\frac{{\sqrt b - 1}}{{\sqrt a + 1}}} = \sqrt {\frac{{\sqrt a - 1}}{{\sqrt b + 1}}.} \sqrt {\frac{{\sqrt a + 1}}{{\sqrt b - 1}}} = \sqrt {\frac{{\sqrt a - 1}}{{\sqrt b + 1}}.\frac{{\sqrt a + 1}}{{\sqrt b - 1}}} \)\( = \sqrt {\frac{{a - 1}}{{b - 1}}} \).