Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\sqrt 2 \\
b,\,\,\,\,\, - \sqrt 5 \\
c,\,\,\,\,\dfrac{{\sqrt 6 }}{2}\\
d,\,\,\,\, - \sqrt a ;\,\,\,\,\sqrt p ;\,\,\,\,\,\sqrt x \\
e,\,\,\,\,\,\sqrt 3 ;\,\,\,\,\dfrac{{\sqrt {14} }}{2};\,\,\,\,\,\sqrt 5
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{2 + \sqrt 2 }}{{1 + \sqrt 2 }} = \dfrac{{{{\sqrt 2 }^2} + \sqrt 2 }}{{1 + \sqrt 2 }} = \dfrac{{\sqrt 2 .\left( {\sqrt 2 + 1} \right)}}{{1 + \sqrt 2 }} = \sqrt 2 \\
b,\\
\dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }} = \dfrac{{\sqrt 5 .\sqrt 3 - \sqrt 5 }}{{1 - \sqrt 3 }} = \dfrac{{\sqrt 5 .\left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }} = - \sqrt 5 \\
c,\\
\dfrac{{2\sqrt 3 - \sqrt 6 }}{{\sqrt 8 - 2}} = \dfrac{{{{\sqrt 2 }^2}.\sqrt 3 - \sqrt 2 .\sqrt 3 }}{{\sqrt {{2^2}.2} - 2}} = \dfrac{{\sqrt 2 .\sqrt 3 .\left( {\sqrt 2 - 1} \right)}}{{2\sqrt 2 - 2}}\\
= \dfrac{{\sqrt 6 .\left( {\sqrt 2 - 1} \right)}}{{2.\left( {\sqrt 2 - 1} \right)}} = \dfrac{{\sqrt 6 }}{2}\\
d,\\
\dfrac{{a - \sqrt a }}{{1 - \sqrt a }} = \dfrac{{{{\sqrt a }^2} - \sqrt a }}{{1 - \sqrt a }} = \dfrac{{\sqrt a .\left( {\sqrt a - 1} \right)}}{{1 - \sqrt a }} = - \sqrt a \\
\dfrac{{p - 2\sqrt p }}{{\sqrt p - 2}} = \dfrac{{{{\sqrt p }^2} - 2\sqrt p }}{{\sqrt p - 2}} = \dfrac{{\sqrt p .\left( {\sqrt p - 2} \right)}}{{\sqrt p - 2}} = \sqrt p \\
\dfrac{{x - \sqrt {xy} }}{{\sqrt x - \sqrt y }} = \dfrac{{{{\sqrt x }^2} - \sqrt {xy} }}{{\sqrt x - \sqrt y }} = \dfrac{{\sqrt x .\left( {\sqrt x - \sqrt y } \right)}}{{\sqrt x - \sqrt y }} = \sqrt x \\
e,\\
\dfrac{{3 + \sqrt 3 }}{{1 + \sqrt 3 }} = \dfrac{{{{\sqrt 3 }^2} + \sqrt 3 }}{{1 + \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{1 + \sqrt 3 }} = \sqrt 3 \\
\dfrac{{\sqrt {14} + \sqrt 7 }}{{2 + \sqrt 2 }} = \dfrac{{\sqrt 7 .\sqrt 2 + \sqrt 7 }}{{{{\sqrt 2 }^2} + \sqrt 2 }} = \dfrac{{\sqrt 7 .\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}} = \dfrac{{\sqrt 7 }}{{\sqrt 2 }} = \dfrac{{\sqrt {14} }}{2}\\
\dfrac{{\sqrt {30} - \sqrt {15} }}{{\sqrt 6 - \sqrt 3 }} = \dfrac{{\sqrt {15} .\sqrt 2 - \sqrt {15} }}{{\sqrt 3 .\sqrt 2 - \sqrt 3 }} = \dfrac{{\sqrt {15} \left( {\sqrt 2 - 1} \right)}}{{\sqrt 3 .\left( {\sqrt 2 - 1} \right)}} = \dfrac{{\sqrt {15} }}{{\sqrt 3 }} = \dfrac{{\sqrt 5 .\sqrt 3 }}{{\sqrt 3 }} = \sqrt 5
\end{array}\)
