Rút gọn các biểu thức sau:

a). \(\sqrt {\frac{{2a}}{5}} .\sqrt {\frac{{5a}}{{18}}} \) với \(a \ge 0\);             

b). \(\sqrt {11a} .\sqrt {\frac{{99}}{a}} \) với \(a > 0\);

c). \(21a - \sqrt {11a} .\sqrt {44a} \) với \(a \ge 0\);                               

d). \({\left( {4 + a} \right)^2} - \sqrt {0,4} .\sqrt {160{a^2}} \)

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

Lời giải

a) Ta có: \(\sqrt {0,49{a^2}}  = 0,7\left| a \right| =  - 0,7a\) (do \(a \le 0\)).

b) Ta có: \[\sqrt {{{\left( {\frac{a}{2}} \right)}^4}{{\left( {6 - 2a} \right)}^2}}  = {\left( {\frac{a}{2}} \right)^2}\left| {6 - 2a} \right| = \frac{{{a^2}.2\left| {3 - a} \right|}}{4} = \frac{{{a^2}\left( {a - 3} \right)}}{2}\] (do \(a > 3\)).

c)  Ta có: \(\sqrt {19.76{{\left( {2 - a} \right)}^2}}  = \sqrt {1444{{\left( {2 - a} \right)}^2}}  = 38\left| {2 - a} \right| = 38\left( {a - 2} \right)\) (do \(a > 2\)).

d) Ta có: \[\frac{1}{{a - b}}.\sqrt {{a^2}{{\left( {{a^2} - {b^2}} \right)}^2}}  = \frac{1}{{a - b}}.\left| a \right|.\left| {{a^2} - {b^2}} \right| = \frac{1}{{a - b}}.a.\left( {a - b} \right)\left( {a + b} \right) = a\left( {a + b} \right)\]

    (do \(a > b \ge 0\)).