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第102頁(yè)

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$解：{\sqrt {5},{\sqrt {\frac {1} {5}}}}是同類(lèi)二次根式，-{\sqrt {10}},{\frac {1} {\sqrt {10}}},{\sqrt {\frac {1} {40}}}是同類(lèi)二次根式$

$解：原式 =(2{\sqrt {6}-2{\sqrt {6}}})+(3{\sqrt {5}-{\sqrt {5}}})$

$=2{\sqrt {5}}$

$解：原式 =({\sqrt {2}+{\frac {\sqrt {2}} {3}}+2{\sqrt {2}}})+({-{\frac {\sqrt {3}} {2}}-{\frac {\sqrt {3}} {5}}}+3{\sqrt {3}})$

$? ={\frac {10} {3}}{\sqrt {2}}+{\frac {23} {10}}{\sqrt {3}}$

$解：原式 ={\sqrt {6}}-2{\sqrt {3}}+3{\sqrt {2}}-2{\sqrt {6}}-4{\sqrt {2}}$

$=({\sqrt {6}-2{\sqrt {6}}})+(3{\sqrt {2}-4{\sqrt {2}}})-2{\sqrt {3}}$

$=-{\sqrt {6}}-{\sqrt {2}}-2{\sqrt {3}}$

$解：原式 ={\sqrt {5}-{\sqrt {10}}+{\frac {\sqrt {10}} {5}}}+{\frac {3{\sqrt {5}}} {5}}-{\frac {\sqrt {10}} {20}}$

$=({\sqrt {5}+{\frac {3{\sqrt {5}}} {5}}})+(-{\sqrt {10}+{\frac {\sqrt {10}} {5}-{\frac {\sqrt {10}} {20}}}})$

$={\frac {8} {5}}{\sqrt {5}}-{\frac {17} {20}}{\sqrt {10}}$

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