Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{3x}}{y}\sqrt {\dfrac{y}{x}} = 3.\sqrt {\dfrac{{{x^2}}}{{{y^2}}}} .\sqrt {\dfrac{y}{x}} = 3.\sqrt {\dfrac{{{x^2}}}{{{y^2}}}.\dfrac{y}{x}} = 3.\sqrt {\dfrac{x}{y}} \\
b,\\
\left( {3x - 1} \right).\sqrt {\dfrac{1}{{3x - 1}}} = \sqrt {{{\left( {3x - 1} \right)}^2}} .\sqrt {\dfrac{1}{{3x - 1}}} = \sqrt {\dfrac{{{{\left( {3x - 1} \right)}^2}}}{{3x - 1}}} = \sqrt {3x - 1} \\
c,\\
\left( {x - 2} \right).\sqrt {\dfrac{1}{{x - 2}}} = \sqrt {{{\left( {x - 2} \right)}^2}} .\sqrt {\dfrac{1}{{x - 2}}} = \sqrt {\dfrac{{{{\left( {x - 2} \right)}^2}}}{{x - 2}}} = \sqrt {x - 2} \\
d,\\
\dfrac{{x + 2\sqrt {xy} + y}}{{\sqrt x + \sqrt y }}.\sqrt {\dfrac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }}} \\
= \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{\sqrt x + \sqrt y }}.\sqrt {\dfrac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }}} \\
= \left( {\sqrt x + \sqrt y } \right).\sqrt {\dfrac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }}} \\
= \sqrt {{{\left( {\sqrt x + \sqrt y } \right)}^2}} .\sqrt {\dfrac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }}} \\
= \sqrt {{{\left( {\sqrt x + \sqrt y } \right)}^2}.\dfrac{{\sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }}} \\
= \sqrt {\left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)} \\
= \sqrt {x - y}
\end{array}\)
