Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
\dfrac{{\sqrt {15} - \sqrt 6 }}{{\sqrt {35} - \sqrt {14} }} = \dfrac{{\sqrt 5 .\sqrt 3 - \sqrt 3 .\sqrt 2 }}{{\sqrt 5 .\sqrt 7 - \sqrt 2 .\sqrt 7 }} = \dfrac{{\sqrt 3 .\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 7 .\left( {\sqrt 5 - \sqrt 2 } \right)}} = \dfrac{{\sqrt 5 }}{{\sqrt 7 }} = \dfrac{{\sqrt {35} }}{7}\\
b,\\
\dfrac{{\sqrt {10} + \sqrt {15} }}{{\sqrt 8 + \sqrt {12} }} = \dfrac{{\sqrt 5 .\sqrt 2 + \sqrt 5 .\sqrt 3 }}{{\sqrt 4 .\sqrt 2 + \sqrt 4 .\sqrt 3 }} = \dfrac{{\sqrt 5 .\left( {\sqrt 2 + \sqrt 3 } \right)}}{{\sqrt 4 .\left( {\sqrt 2 + \sqrt 3 } \right)}} = \dfrac{{\sqrt 5 }}{{\sqrt 4 }} = \dfrac{{\sqrt 5 }}{2}\\
c,\\
\dfrac{{x + \sqrt {xy} }}{{y + \sqrt {xy} }} = \dfrac{{{{\sqrt x }^2} + \sqrt {xy} }}{{{{\sqrt y }^2} + \sqrt {xy} }} = \dfrac{{\sqrt x .\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt y .\left( {\sqrt y + \sqrt x } \right)}} = \dfrac{{\sqrt x }}{{\sqrt y }} = \dfrac{{\sqrt {xy} }}{y}\\
3,\\
a,\\
\sqrt {9{{\left( {3 - a} \right)}^2}} = \sqrt {{3^2}.{{\left( {3 - a} \right)}^2}} = \left| {3\left( {3 - a} \right)} \right| = 3.\left| {3 - a} \right| = 3.\left( {a - 3} \right)\\
\left( {a > 3 \Rightarrow 3 - a < 0 \Rightarrow \left| {3 - a} \right| = - \left( {3 - a} \right) = a - 3} \right)\\
b,\\
\sqrt {{a^2}{{\left( {a - 2} \right)}^2}} = \left| {a.\left( {a - 2} \right)} \right| = a.\left( {a - 2} \right)\\
\left( {a < 0 \Rightarrow \left\{ \begin{array}{l}
a < 0\\
a - 2 < 0
\end{array} \right. \Rightarrow a\left( {a - 2} \right) > 0 \Rightarrow \left| {a.\left( {a - 2} \right)} \right| = a.\left( {a - 2} \right)} \right)
\end{array}\)
