Khử mẫu trong dấu căn:

a) \(2a.\sqrt {\frac{3}{5}} ;\)

b) \( - 3x.\sqrt {\frac{5}{x}} \) (x > 0);

c) \[ - \sqrt {\frac{{3a}}{b}} \] (a ≥ 0, b > 0).

Ta có \[A = \sqrt x \left( {\frac{1}{{\sqrt x + 3}} - \frac{1}{{3 - \sqrt x }}} \right)\]

\[ = \sqrt x \cdot \frac{{3 - \sqrt x - \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {3 - \sqrt x } \right)}}\]

\[ = \sqrt x .\frac{{ - 2\sqrt x }}{{{3^2} - {{\left( {\sqrt x } \right)}^2}}} = \frac{{ - 2x}}{{9 - x}}.\]

a) \(2\sqrt {\frac{2}{3}} - 4\sqrt {\frac{3}{2}} = 2\sqrt {\frac{{2.3}}{{{3^2}}}} - 4\sqrt {\frac{{3.2}}{{{2^2}}}} \)

\( = \frac{2}{3}\sqrt 6 - 2\sqrt 6 = - \frac{4}{3}\sqrt 6 \).

b) \(\frac{{5\sqrt {48} - 3\sqrt {27} + 2\sqrt {12} }}{{\sqrt 3 }}\)

\( = \frac{{5\sqrt {48} }}{{\sqrt 3 }} - \frac{{3\sqrt {27} }}{{\sqrt 3 }} + \frac{{2\sqrt {12} }}{{\sqrt 3 }}\)

\( = 5\sqrt {\frac{{48}}{3}} - 3\sqrt {\frac{{27}}{3}} + 2\sqrt {\frac{{12}}{3}} \)

\( = 5\sqrt {16} - 3\sqrt 9 + 2\sqrt 4 \)

= 5.4 – 3.3 + 2.2 = 20 – 9 + 4 = 15.

c) \[\frac{1}{{3 + 2\sqrt 2 }} + \frac{{4\sqrt 2 - 4}}{{2 - \sqrt 2 }}\]

\[ = \frac{{3 - 2\sqrt 2 }}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}} + \frac{{4\left( {\sqrt 2 - 1} \right)}}{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}\]

\( = \frac{{3 - 2\sqrt 2 }}{{{3^2} - {{\left( {2\sqrt 2 } \right)}^2}}} + \frac{{{{\left( {\sqrt 2 } \right)}^4}}}{{\sqrt 2 }} = \frac{{3 - 2\sqrt 2 }}{{9 - 8}} + {\left( {\sqrt 2 } \right)^3}\)

\( = 3 - 2\sqrt 2 + 2\sqrt 2 = 3.\)