Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x{y^2}.\sqrt {\dfrac{3}{{{x^2}{y^4}}}} = x{y^2}.\sqrt {\dfrac{3}{{{{\left( {x{y^2}} \right)}^2}}}} = x{y^2}.\dfrac{{\sqrt 3 }}{{\left| {x{y^2}} \right|}} = x{y^2}.\dfrac{{\sqrt 3 }}{{x{y^2}}} = \sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {do\,\,x > 0 \Rightarrow x{y^2} > 0} \right)\\
b,\\
x{y^2}.\sqrt {\dfrac{3}{{{x^2}{y^4}}}} = x{y^2}.\sqrt {\dfrac{3}{{{{\left( {x{y^2}} \right)}^2}}}} = x{y^2}.\dfrac{{\sqrt 3 }}{{\left| {x{y^2}} \right|}} = x{y^2}.\dfrac{{\sqrt 3 }}{{ - x{y^2}}} = - \sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {do\,\,x < 0 \Rightarrow x{y^2} < 0} \right)\\
c,\\
\sqrt {\dfrac{{27{{\left( {x - 4} \right)}^2}}}{{48}}} = \sqrt {\dfrac{{27}}{{48}}.{{\left( {x - 4} \right)}^2}} = \sqrt {\dfrac{9}{{16}}.{{\left( {x - 4} \right)}^2}} = \dfrac{3}{4}.\left| {x - 4} \right| = \dfrac{3}{4}\left( {x - 4} \right)\,\,\,\,\,\,\,\,\,\left( {do\,\,\,x > 4 \Rightarrow x - 4 > 0} \right)\\
d,\\
\sqrt {\dfrac{{27{{\left( {x - 4} \right)}^2}}}{{48}}} = \sqrt {\dfrac{{27}}{{48}}.{{\left( {x - 4} \right)}^2}} = \sqrt {\dfrac{9}{{16}}.{{\left( {x - 4} \right)}^2}} = \dfrac{3}{4}.\left| {x - 4} \right| = - \dfrac{3}{4}\left( {x - 4} \right)\,\,\,\,\,\,\,\,\,\left( {do\,\,\,x < 4 \Rightarrow x - 4 < 0} \right)
\end{array}\)
