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第101頁

信息發(fā)布者：

$? ={\frac {1×{\sqrt {2}}} {\sqrt {2}×{\sqrt {2}}}}$

$={\frac {\sqrt {2}} {2}}$

$?={\frac {1×{\sqrt {3}}} {\sqrt {3}×{\sqrt {3}}}}$

$={\frac {\sqrt {3}} {3}}$

$={\frac {\sqrt {2}×{\sqrt {5}}} {\sqrt {5}×{\sqrt {5}}}}$

$? ={\frac {\sqrt {10}} {5}}$

$={\frac {\sqrt {5}×{\sqrt {6}}} {\sqrt {6}×{\sqrt {6}}}}$

$? ={\frac {\sqrt {30}} {6}}$

$={\frac {\sqrt {3}×{\sqrt {2}}} {2{\sqrt {2}}×{\sqrt {2}}}}$

$={\frac {\sqrt {6}} {4}}$

$?={\frac {1×{\sqrt {3}}} {2{\sqrt {3}}×{\sqrt {3}}}}$

$={\frac {\sqrt {3}} {6}}$

$={\frac {\sqrt {7}×{\sqrt {2}}} {3{\sqrt {2}×{\sqrt {2}}}}}$

$={\frac {\sqrt {14}} {6}}$

$?={\frac {\sqrt {5}×{\sqrt {3}}} {3{\sqrt {3}×{\sqrt {3}}}}}$

$={\frac {\sqrt {15}} {9}}$

$={\frac {{\sqrt {y}×}\sqrt {2x}} {\sqrt {2x}×{\sqrt {2x}}}}$

$={\frac {\sqrt {2xy}} {2x}}$

$?解： {\sqrt {2}},{\frac {\sqrt {2}} {3}},-2{\sqrt {2}}是同類二次根式，{\frac {\sqrt {3}} {2}},{\frac {\sqrt {3}} {5}},3{\sqrt {3}}是同類二次根式$

$解：{\sqrt {6},{\sqrt {24}}}是同類二次根式，{\sqrt {18},{\sqrt {32}}}是同類二次根式$

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