Rút gọn các biểu thức sau:

    a) \[\frac{y}{x}\sqrt {\frac{{{x^2}}}{{{y^4}}}} \] với \[x > 0,y \ne 0\].                    b)\[2{y^2}\sqrt {\frac{{{x^4}}}{{4{y^2}}}} \] với \[y < 0\].

    c) \[5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} \]với \[x < 0,y > 0\].                                d) \[0,2{x^3}{y^3}\sqrt {\frac{{16}}{{{x^4}{y^8}}}} \]với \[x \ne 0,y \ne 0\].

b)  Ta có: \(\sqrt {11a} .\sqrt {\frac{{99}}{a}}  = \sqrt {11a.\frac{{99}}{a}}  = \sqrt {1089}  = 33\).

c)  Ta có: \(21a - \sqrt {11a} .\sqrt {44a}  = 21a - \sqrt {484{a^2}}  = 21a - 22a =  - a\) (do \(a \ge 0\))

d) Ta có: \({\left( {4 + a} \right)^2} - \sqrt {0,4} .\sqrt {160{a^2}}  = {\left( {4 + a} \right)^2} - \sqrt {64{a^2}}  = {\left( {4 + a} \right)^2} - 8\left| a \right| = 16 + {a^2} + 8a - 8\left| a \right|\)

 Nếu \(a \ge 0\) thì \({\left( {4 + a} \right)^2} - \sqrt {0,4} .\sqrt {160{a^2}}  = 16 + {a^2}\).

 Nếu \(a < 0\) thì \({\left( {4 + a} \right)^2} - \sqrt {0,4} .\sqrt {160{a^2}}  = 16 + {a^2} + 16a\).

a) Ta có: \[a{b^2}\sqrt {\frac{3}{{{a^2}{b^4}}}}  = a{b^2}.\frac{{\sqrt 3 }}{{\left| {a{b^2}} \right|}} = a{b^2}.\frac{{\sqrt 3 }}{{ - a{b^2}}} =  - \sqrt 3 \] ( do \[a < 0,b \ne 0\]).

b) Ta có \[\sqrt {\frac{{27{{\left( {a - 3} \right)}^2}}}{{48}}}  = \sqrt {\frac{{9{{\left( {a - 3} \right)}^2}}}{{16}}}  = \frac{{3\left| {a - 3} \right|}}{4} = \frac{{3\left( {a - 3} \right)}}{4}\] (do \[a > 3\]).

c)  Ta có: \[\sqrt {\frac{{9 + 12a + 4{a^2}}}{{{b^2}}}}  = \sqrt {\frac{{{{\left( {3 + 2a} \right)}^2}}}{{{b^2}}}}  = \frac{{\left| {3 + 2a} \right|}}{{\left| b \right|}} = \frac{{3 + 2a}}{{ - b}}\]. (do\[b < 0,a >  - 1,5\]).

d) \[\left( {a - b} \right)\sqrt {\frac{{ab}}{{{{\left( {a - b} \right)}^2}}}}  = \left( {a - b} \right)\frac{{\sqrt {ab} }}{{\sqrt {{{\left( {a - b} \right)}^2}} }} = \left( {a - b} \right)\frac{{\sqrt {ab} }}{{\left| {a - b} \right|}} = \left( {a - b} \right)\frac{{\sqrt {ab} }}{{b - a}} =  - \sqrt {ab} \] (do \[a < b < 0)\]