Rút gọn biểu thức sau:

a). \(\sqrt {0,49{a^2}} \) với \(a \le 0\);                                                                       

b). \(\sqrt {{{\left( {\frac{a}{2}} \right)}^4}{{\left( {6 - 2a} \right)}^2}} \) với \(a > 3\);

c). \(\sqrt {19.76{{\left( {2 - a} \right)}^2}} \) với \(a > 2\);    

d). \[\frac{1}{{a - b}}.\sqrt {{a^2}{{\left( {{a^2} - {b^2}} \right)}^2}} \] với \(a > b \ge 0\).

a) Ta có: \[\frac{y}{x}\sqrt {\frac{{{x^2}}}{{{y^4}}}}  = \frac{y}{x}.\frac{{\left| x \right|}}{{{y^2}}} = \frac{y}{x}.\frac{x}{{{y^2}}} = \frac{1}{y}\]( do \[x > 0,y \ne 0\]).

b) Ta có \[2{y^2}\sqrt {\frac{{{x^4}}}{{4{y^2}}}}  = 2{y^2}\sqrt {\frac{{{x^4}}}{{4{y^2}}}}  = 2{y^2}\frac{{{x^2}}}{{2\left| y \right|}} = 2{y^2}.\frac{{{x^2}}}{{ - 2y}} =  - {x^2}y\].

c)  Ta có:. \[5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}}  = 5xy.\frac{{\sqrt {25{x^2}} }}{{\sqrt {{y^6}} }} = 5xy.\frac{{5\left| x \right|}}{{\left| {{y^3}} \right|}} = 5xy.\frac{{ - 5x}}{{{y^3}}} =  - \frac{{25{x^2}}}{{{y^2}}}\] (do \[x < 0,y > 0\]).

d) \[0,2{x^3}{y^3}\sqrt {\frac{{16}}{{{x^4}{y^8}}}}  = 0,2{x^3}{y^3}.\frac{4}{{{x^2}{y^4}}} = \frac{{0,8x}}{y}.\]

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 \[a < b < 0)\]