Đưa thừa số ra ngoài dấu căn:

a). \[\sqrt {96.125} \].                                                                    b). \[\sqrt {{a^4}{b^5}} \].

c). \[\sqrt {{a^6}{b^{11}}} \].                                                                    d). \[\sqrt {{a^3}{{\left( {1 - a} \right)}^4}} \left( {a > 1} \right)\].

a) \(\left( {\sqrt {2x}  - 1} \right)\left( {2x + \sqrt {2x}  + 1} \right) = \left( {\sqrt {2x}  - 1} \right)\left( {{{\left( {\sqrt {2x} } \right)}^2} + \sqrt {2x} .1 + {1^2}} \right) = {\left( {\sqrt {2x} } \right)^3} - 1 = 2x\sqrt 2 x - 1\).

b) \(\left( {\sqrt x  + 2\sqrt y } \right)\left( {x - 2\sqrt {xy}  + 4y} \right) = \left( {\sqrt x  + 2\sqrt y } \right)\left[ {{{\left( {\sqrt x } \right)}^2} - \sqrt x .2\sqrt y  + {{\left( {2\sqrt y } \right)}^2}} \right]\)

\( = {\left( {\sqrt x } \right)^3} + {\left( {2\sqrt y } \right)^3} = x\sqrt x  + 8y\sqrt y \).

c) \(\left( {\sqrt x  + \sqrt y } \right)\left( {2\sqrt x  - \sqrt y } \right) = \sqrt x .2\sqrt x  - \sqrt x .\sqrt y  + \sqrt y .2\sqrt x  - \sqrt y .\sqrt y \)

\( = 2x - \sqrt {xy}  + 2\sqrt {xy}  - y = 2x + \sqrt {xy}  - y\).

a). Với \(a \ge 0\), ta có: \(3\sqrt {2a}  - \sqrt {18{a^3}}  + 4\sqrt {\frac{a}{2}}  - \frac{1}{4}\sqrt {128a} \)

\( = 3\sqrt {2a}  - 3a\sqrt {2a}  + 2\sqrt {2a}  - 2\sqrt {2a} \)

\( = 3\sqrt {2a}  - 3a\sqrt {2a}  = 3\left( {1 - a} \right)\sqrt {2a} .\)

b). Với \(x > y > 0\), ta có: \(2y\sqrt {x - y}  + x\sqrt {\frac{1}{{x - y}}}  - x\sqrt {\frac{a}{{ax - ay}}}  - \sqrt {{x^3} - {x^2}y} \)

\( = 2y\sqrt {x - y}  + x\sqrt {\frac{1}{{x - y}}}  - x\sqrt {\frac{a}{{a\left( {x - y} \right)}}}  - \sqrt {{x^2}\left( {x - y} \right)} \)

\( = 2y\sqrt {x - y}  - \left| x \right|\sqrt {x - y}  + x\sqrt {\frac{1}{{x - y}}}  - x\sqrt {\frac{1}{{x - y}}} \)

\( = \sqrt {x - y} \left( {2y - x} \right)\) (do \(x > 0\)).

c). Với \(a \ge 0,b \ge 0,a \ne b\), ta có:
\(\frac{{\sqrt a  + \sqrt b }}{{\sqrt a  - \sqrt b }} + \frac{{\sqrt a  - \sqrt b }}{{\sqrt a  + \sqrt b }} = \frac{{{{\left( {\sqrt a  + \sqrt b } \right)}^2} + {{\left( {\sqrt a  - \sqrt b } \right)}^2}}}{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}\)

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