Đáp án:
\(\begin{array}{l}
c,\,\,\,\,\,3\sqrt 7 \\
d,\,\,\,\,\dfrac{3}{2}\sqrt 5 \\
e,\,\,\,\,\,5 + 2\sqrt 6
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\\
42\sqrt {\dfrac{1}{7}} - \sqrt {112} + \dfrac{{\sqrt 7 + 7}}{{\sqrt 7 + 1}}\\
= 42\sqrt {\dfrac{1}{{49}}.7} - \sqrt {16.7} + \dfrac{{\sqrt 7 + {{\sqrt 7 }^2}}}{{\sqrt 7 + 1}}\\
= 42\sqrt {{{\left( {\dfrac{1}{7}} \right)}^2}.7} - \sqrt {{4^2}.7} + \dfrac{{\sqrt 7 .\left( {1 + \sqrt 7 } \right)}}{{\sqrt 7 + 1}}\\
= 42.\dfrac{1}{7}\sqrt 7 - 4\sqrt 7 + \sqrt 7 \\
= 6\sqrt 7 - 4\sqrt 7 + \sqrt 7 \\
= 3\sqrt 7 \\
d,\\
\dfrac{2}{{\sqrt 7 - \sqrt 5 }} - \dfrac{1}{3}\sqrt {63} + \dfrac{1}{4}\sqrt {20} \\
= \dfrac{{2.\left( {\sqrt 7 + \sqrt 5 } \right)}}{{\left( {\sqrt 7 - \sqrt 5 } \right)\left( {\sqrt 7 + \sqrt 5 } \right)}} - \dfrac{1}{3}\sqrt {{3^2}.7} + \dfrac{1}{4}\sqrt {{2^2}.5} \\
= \dfrac{{2.\left( {\sqrt 7 + \sqrt 5 } \right)}}{{{{\sqrt 7 }^2} - {{\sqrt 5 }^2}}} - \dfrac{1}{3}.3\sqrt 7 + \dfrac{1}{4}.2\sqrt 5 \\
= \dfrac{{2.\left( {\sqrt 7 + \sqrt 5 } \right)}}{2} - \sqrt 7 + \dfrac{1}{2}\sqrt 5 \\
= \sqrt 7 + \sqrt 5 - \sqrt 7 + \dfrac{1}{2}\sqrt 5 \\
= \dfrac{3}{2}\sqrt 5 \\
e,\\
\dfrac{{3\sqrt 3 - 2\sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + 3\sqrt {\dfrac{2}{3}} \\
= \dfrac{{{{\sqrt 3 }^3} - {{\sqrt 2 }^3}}}{{\sqrt 3 - \sqrt 2 }} + \sqrt {{3^2}} .\sqrt {\dfrac{2}{3}} \\
= \dfrac{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {{{\sqrt 3 }^2} + \sqrt 3 .\sqrt 2 + {{\sqrt 2 }^2}} \right)}}{{\sqrt 3 - \sqrt 2 }} + \sqrt {{3^2}.\dfrac{2}{3}} \\
= \dfrac{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {3 + \sqrt 6 + 2} \right)}}{{\sqrt 3 - \sqrt 2 }} + \sqrt 6 \\
= \left( {3 + \sqrt 6 + 2} \right) + \sqrt 6 \\
= 5 + 2\sqrt 6
\end{array}\)
