Gấp dùm mình ạ, cảm ơn nhìuuuuuuuu

Đáp án:

$\begin{array}{l}
7A)\\
a)\left( {x \ge 0;y > 0} \right)\\
\sqrt {\dfrac{{5{x^2}}}{{49y}}}  = \dfrac{{\sqrt 5 .\left| x \right|}}{{7\sqrt y }} = \dfrac{{\sqrt 5 .x.\sqrt y }}{{7y}}\\
b)x < 0;y > 0\\
7xy\sqrt {\dfrac{{ - 3}}{{xy}}} \\
 = 7xy.\sqrt {\dfrac{{ - 3xy}}{{{x^2}{y^2}}}} \\
 = 7xy.\dfrac{{\sqrt { - 3xy} }}{{ - xy}}\\
 =  - 7\sqrt { - 3xy} \\
7B)\\
a)a > 0;b \ge 0\\
\sqrt {\dfrac{{5b}}{{49{a^2}}}}  = \dfrac{{\sqrt {5b} }}{{7a}}\\
b) - \dfrac{1}{4}ab\sqrt {\dfrac{{16}}{{ab}}} \left( {a < 0;b < 0} \right)\\
 =  - \dfrac{1}{4}ab.\dfrac{4}{{\sqrt {ab} }}\\
 =  - \sqrt {ab} 
\end{array}$