Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{\sqrt {15} - \sqrt 6 }}{{\sqrt {35} - \sqrt {14} }} = \dfrac{{\sqrt 5 .\sqrt 3 - \sqrt 3 .\sqrt 2 }}{{\sqrt 5 .\sqrt 7 - \sqrt 7 .\sqrt 2 }} = \dfrac{{\sqrt 3 .\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 7 \left( {\sqrt 5 - \sqrt 2 } \right)}} = \dfrac{{\sqrt 3 }}{{\sqrt 7 }} = \dfrac{{\sqrt {21} }}{7}\\
b,\\
\dfrac{{\sqrt {10} + \sqrt {15} }}{{\sqrt 8 + \sqrt {12} }} = \dfrac{{\sqrt 2 .\sqrt 5 + \sqrt 5 .\sqrt 3 }}{{\sqrt 2 .\sqrt 4 + \sqrt 3 .\sqrt 4 }} = \dfrac{{\sqrt 5 .\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 4 .\left( {\sqrt 3 + \sqrt 2 } \right)}} = \dfrac{{\sqrt 5 }}{{\sqrt 4 }} = \dfrac{{\sqrt 5 }}{2}\\
c,\\
\dfrac{{2\sqrt {15} - 2\sqrt {10} + \sqrt 6 - 3}}{{2\sqrt 5 - 2\sqrt {10} - \sqrt 3 + \sqrt 6 }}\\
= \dfrac{{2\sqrt 5 .\sqrt 3 - 2.\sqrt 5 .\sqrt 2 + \sqrt 3 .\sqrt 2 - {{\sqrt 3 }^2}}}{{2\sqrt 5 - 2.\sqrt 5 .\sqrt 2 - \sqrt 3 + \sqrt 3 .\sqrt 2 }}\\
= \dfrac{{2\sqrt 5 .\left( {\sqrt 3 - \sqrt 2 } \right) + \sqrt 3 .\left( {\sqrt 2 - \sqrt 3 } \right)}}{{2\sqrt 5 \left( {1 - \sqrt 2 } \right) - \sqrt 3 \left( {1 - \sqrt 2 } \right)}}\\
= \dfrac{{\left( {\sqrt 3 - \sqrt 2 } \right).\left( {2\sqrt 5 - \sqrt 3 } \right)}}{{\left( {1 - \sqrt 2 } \right).\left( {2\sqrt 5 - \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 }}{{1 - \sqrt 2 }}\\
d,\\
\dfrac{{\sqrt 2 + \sqrt 3 + \sqrt 6 + \sqrt 8 + \sqrt {16} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}\\
= \dfrac{{\sqrt 2 + \sqrt 3 + \sqrt 6 + 2\sqrt 2 + 4}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= \dfrac{{\left( {\sqrt 2 + \sqrt 3 + 2} \right) + \left( {\sqrt 6 + 2\sqrt 2 + 2} \right)}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= \dfrac{{\left( {\sqrt 2 + \sqrt 3 + 2} \right) + \sqrt 2 .\left( {\sqrt 3 + 2 + \sqrt 2 } \right)}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= \dfrac{{\left( {\sqrt 2 + \sqrt 3 + 2} \right).\left( {1 + \sqrt 2 } \right)}}{{\sqrt 2 + \sqrt 3 + 2}}\\
= 1 + \sqrt 2 \\
e,\\
\dfrac{{x + \sqrt {xy} }}{{y + \sqrt {xy} }} = \dfrac{{\sqrt x \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt y \left( {\sqrt y + \sqrt x } \right)}} = \dfrac{{\sqrt x }}{{\sqrt y }}\\
f,\\
\dfrac{{\sqrt a + a\sqrt b - \sqrt b - b\sqrt a }}{{ab - 1}}\\
= \dfrac{{\sqrt a \left( {1 + \sqrt {ab} } \right) - \sqrt b .\left( {1 + \sqrt {ab} } \right)}}{{\left( {\sqrt {ab} - 1} \right)\left( {\sqrt {ab} + 1} \right)}}\\
= \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {1 + \sqrt {ab} } \right)}}{{\left( {\sqrt {ab} - 1} \right)\left( {\sqrt {ab} + 1} \right)}}\\
= \dfrac{{\sqrt a - \sqrt b }}{{\sqrt {ab} - 1}}
\end{array}\)
