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第97頁(yè)

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$解：\sqrt {12{a}^{2}^{3}}$
$={\sqrt {4{a}^{2}^{2}·3b}}$
$={\sqrt {4{a}^{2}^{2}}}·{\sqrt {3b}}$
$=2ab{\sqrt {3b}}$
$解：\sqrt {{13}^{2}-{12}^{2}}$
$={\sqrt {169-144}}$
$={\sqrt {25}}$
$=5$
$解：{\sqrt {120}}$
$={\sqrt {4×30}}$
$={\sqrt {4}}×{\sqrt {30}}$
$=2{\sqrt {30}}$
$解：{\sqrt {200}}$
$={\sqrt {100×2}}$
$={\sqrt {100}}×{\sqrt {2}}$
$=10{\sqrt {2}}$
$解：{\sqrt {8{x}^{3}y}}$
$={\sqrt {4{x}^{2}·2xy}}$
$={\sqrt {4{x}^{2}}}·{\sqrt {2xy}}$
$=2x{\sqrt {2xy}}$
$解：{\sqrt {12{x}^{3}{y}^{5}}}$
$={\sqrt {4{x}^{2}{y}^{4}·3xy}}$
$={\sqrt {4{x}^{2}{y}^{4}}·{\sqrt {3xy}}}$
$=2x{y}^{2}{\sqrt {3xy}}$
$解：原式 ={\sqrt {{x}^{4}(x+1)}}$
$={\sqrt {{x}^{4}}·}{\sqrt {x+1}}$
$={x}^{2}{\sqrt {x+1}}$
$?解：原式 ={\sqrt {a({a}^{4}+2{a}^{2}^{2}+^{4})}}$
$={\sqrt {a{({a}^{2}+^{2})}^{2}}}$
$={\sqrt {a}}·{\sqrt {{({a}^{2}+^{2})}^{2}}}$
$=({a}^{2}+^{2}){\sqrt {a}}$
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