Đáp án:
\(\begin{array}{l}
a,\,\,\,\, - \sqrt 3 \\
b,\,\,\,\,\,3\sqrt 6 - 6\\
c,\,\,\,\,\, - \sqrt 5 \\
d,\,\,\,\,\,\dfrac{{\sqrt 6 }}{2}\\
e,\,\,\,\,\,\sqrt 3 + 1\\
f,\,\,\,\,\,\dfrac{{2 + \sqrt 2 - \sqrt 6 }}{4}\\
g,\,\,\,\, - 8 + 3\sqrt 7 \\
h,\,\,\,\,a + \sqrt {ab} + b\\
i,\,\,\,\,\,4\sqrt 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {75} + \sqrt {48} - \sqrt {300} \\
= \sqrt {25.3} + \sqrt {16.3} - \sqrt {100.3} \\
= \sqrt {{5^2}.3} + \sqrt {{4^2}.3} - \sqrt {{{10}^2}.3} \\
= 5\sqrt 3 + 4\sqrt 3 - 10\sqrt 3 \\
= - \sqrt 3 \\
b,\\
\sqrt {18{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} \\
= \sqrt {{3^2}.2{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} \\
= \sqrt {{{\left( {3\sqrt 2 } \right)}^2}.{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} \\
= 3\sqrt 2 .\left| {\sqrt 2 - \sqrt 3 } \right|\\
= 3\sqrt 2 .\left( {\sqrt 3 - \sqrt 2 } \right)\\
= 3.\sqrt 2 .\sqrt 3 - 3{\sqrt 2 ^2}\\
= 3\sqrt {3.2} - 3.2\\
= 3\sqrt 6 - 6\\
c,\\
\dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }} = \dfrac{{\sqrt {5.3} - \sqrt 5 }}{{1 - \sqrt 3 }} = \dfrac{{\sqrt 5 .\left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }} = - \sqrt 5 \\
d,\\
\dfrac{{9 - 2\sqrt 3 }}{{3\sqrt 6 - 2\sqrt 2 }} = \dfrac{{3.3 - 2\sqrt 3 }}{{3\sqrt {3.2} - 2\sqrt 2 }} = \dfrac{{3.{{\sqrt 3 }^2} - 2\sqrt 3 }}{{3\sqrt 3 .\sqrt 2 - 2\sqrt 2 }}\\
= \dfrac{{\sqrt 3 .\left( {3\sqrt 3 - 2} \right)}}{{\sqrt 2 .\left( {3\sqrt 3 - 2} \right)}} = \dfrac{{\sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt 3 .\sqrt 2 }}{{{{\sqrt 2 }^2}}} = \dfrac{{\sqrt 6 }}{2}\\
e,\\
\dfrac{2}{{\sqrt 3 - 1}} = \dfrac{{2.\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} = \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{{{\sqrt 3 }^2} - {1^2}}}\\
= \dfrac{{2.\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} = \dfrac{{2.\left( {\sqrt 3 + 1} \right)}}{2} = \sqrt 3 + 1\\
f,\\
\dfrac{1}{{\sqrt 3 + \sqrt 2 + 1}} = \dfrac{1}{{\sqrt 3 + \left( {\sqrt 2 + 1} \right)}}\\
= \dfrac{{\sqrt 3 - \left( {\sqrt 2 + 1} \right)}}{{\left[ {\sqrt 3 - \left( {\sqrt 2 + 1} \right)} \right].\left[ {\sqrt 3 + \left( {\sqrt 2 + 1} \right)} \right]}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 - 1}}{{{{\sqrt 3 }^2} - {{\left( {\sqrt 2 + 1} \right)}^2}}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 - 1}}{{3 - \left( {{{\sqrt 2 }^2} + 2.\sqrt 2 .1 + {1^2}} \right)}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 - 1}}{{3 - \left( {2 + 2\sqrt 2 + 1} \right)}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 - 1}}{{3 - \left( {3 + 2\sqrt 2 } \right)}}\\
= \dfrac{{\sqrt 3 - \sqrt 2 - 1}}{{ - 2\sqrt 2 }}\\
= \dfrac{{\sqrt 2 .\left( {\sqrt 3 - \sqrt 2 - 1} \right)}}{{ - 2.{{\sqrt 2 }^2}}}\\
= \dfrac{{\sqrt 6 - {{\sqrt 2 }^2} - \sqrt 2 }}{{ - 2.2}}\\
= \dfrac{{\sqrt 6 - 2 - \sqrt 2 }}{{ - 4}}\\
= \dfrac{{2 + \sqrt 2 - \sqrt 6 }}{4}\\
g,\\
\dfrac{{\sqrt 7 - 3}}{{\sqrt 7 + 3}} = \dfrac{{{{\left( {\sqrt 7 - 3} \right)}^2}}}{{\left( {\sqrt 7 + 3} \right)\left( {\sqrt 7 - 3} \right)}}\\
= \dfrac{{{{\sqrt 7 }^2} - 2.\sqrt 7 .3 + {3^2}}}{{{{\sqrt 7 }^2} - {3^2}}} = \dfrac{{7 - 6\sqrt 7 + 9}}{{7 - 9}}\\
= \dfrac{{16 - 6\sqrt 7 }}{{ - 2}} = - 8 + 3\sqrt 7 \\
h,\\
\dfrac{{a\sqrt a - b\sqrt b }}{{\sqrt a - \sqrt b }} = \dfrac{{{{\sqrt a }^3} - {{\sqrt b }^3}}}{{\sqrt a - \sqrt b }}\\
= \dfrac{{\left( {\sqrt a - \sqrt b } \right).\left( {{{\sqrt a }^2} + \sqrt a .\sqrt b + {{\sqrt b }^2}} \right)}}{{\sqrt a - \sqrt b }}\\
= \dfrac{{\left( {\sqrt a - \sqrt b } \right).\left( {a + \sqrt {ab} + b} \right)}}{{\sqrt a - \sqrt b }}\\
= a + \sqrt {ab} + b\\
i,\\
3\sqrt 8 - 4\sqrt {18} + 2\sqrt {50} \\
= 3\sqrt {4.2} - 4\sqrt {9.2} + 2\sqrt {25.2} \\
= 3\sqrt {{2^2}.2} - 4\sqrt {{3^2}.2} + 2\sqrt {{5^2}.2} \\
= 3.2\sqrt 2 - 4.3\sqrt 2 + 2.5\sqrt 2 \\
= 6\sqrt 2 - 12\sqrt 2 + 10\sqrt 2 \\
= 4\sqrt 2
\end{array}\)
