Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\sqrt {12} + 2\sqrt {27} + 3\sqrt {75} - 9\sqrt {48} \\
= \sqrt {{2^2}.3} + 2.\sqrt {{3^2}.3} + 3\sqrt {{5^2}.3} - 9.\sqrt {{4^2}.3} \\
= 2\sqrt 3 + 2.3\sqrt 3 + 3.5\sqrt 3 - 9.4\sqrt 3 \\
= 2\sqrt 3 + 6\sqrt 3 + 15\sqrt 3 - 36\sqrt 3 \\
= - 13\sqrt 3 \\
b,\\
2\sqrt 2 - {\sqrt 3 ^2} = 2\sqrt 2 - 3\\
c,\\
\left( {\sqrt x - 3} \right)\left( {4 - \sqrt x } \right) = 4\sqrt x - {\sqrt x ^2} - 12 + 3\sqrt x = - {x^2} + 7\sqrt x - 12\\
2,\\
a,\\
\dfrac{{\sqrt {15} - \sqrt 6 }}{{\sqrt {35} - \sqrt {14} }} = \dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 7 \left( {\sqrt 5 - \sqrt 2 } \right)}} = \dfrac{{\sqrt 3 }}{{\sqrt 7 }} = \dfrac{{\sqrt {21} }}{7}\\
b,\\
\dfrac{{\sqrt {10} + \sqrt {15} }}{{\sqrt 8 + \sqrt {12} }} = \dfrac{{\sqrt 5 \left( {\sqrt 2 + \sqrt 3 } \right)}}{{\sqrt 4 \left( {\sqrt 2 + \sqrt 3 } \right)}} = \dfrac{{\sqrt 5 }}{{\sqrt 4 }} = \dfrac{{\sqrt 5 }}{2}\\
c,\\
\dfrac{{2\sqrt {15} - 2\sqrt {10} + \sqrt 6 - 3}}{{2\sqrt 5 - 2\sqrt {10} - \sqrt 3 + \sqrt 6 }}\\
= \dfrac{{\left( {2\sqrt {15} - 2\sqrt {10} } \right) + \left( {\sqrt 6 - 3} \right)}}{{\left( {2\sqrt 5 - \sqrt 3 } \right) - \left( {2\sqrt {10} - \sqrt 6 } \right)}}\\
= \dfrac{{2\sqrt 5 \left( {\sqrt 3 - \sqrt 2 } \right) + \sqrt 3 \left( {\sqrt 2 - \sqrt 3 } \right)}}{{\left( {2\sqrt 5 - \sqrt 3 } \right) - \sqrt 2 \left( {2\sqrt 5 - \sqrt 3 } \right)}}\\
= \dfrac{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {2\sqrt 5 - \sqrt 3 } \right)}}{{\left( {2\sqrt 5 - \sqrt 3 } \right)\left( {1 - \sqrt 2 } \right)}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 }}{{1 - \sqrt 2 }}\\
d,\\
\dfrac{{x - 2\sqrt x }}{{x + 4 - 4\sqrt x }} = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x - 4\sqrt x + 4}} = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{{{\left( {\sqrt x - 2} \right)}^2}}} = \dfrac{{\sqrt x }}{{\sqrt x - 2}}
\end{array}\)
