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$解：\sqrt {12{a}^{2}{b}^{3}}$

$={\sqrt {4{a}^{2}{b}^{2}·3b}}$

$={\sqrt {4{a}^{2}{b}^{2}}}·{\sqrt {3b}}$

$=2ab{\sqrt {3b}}$

$解：\sqrt {{13}^{2}-{12}^{2}}$

$={\sqrt {169-144}}$

$={\sqrt {25}}$

$=5$

$解：{\sqrt {120}}$

$={\sqrt {4×30}}$

$={\sqrt {4}}×{\sqrt {30}}$

$=2{\sqrt {30}}$

$解：{\sqrt {200}}$

$={\sqrt {100×2}}$

$={\sqrt {100}}×{\sqrt {2}}$

$=10{\sqrt {2}}$

$解：{\sqrt {12{x}^{3}{y}^{5}}}$

$={\sqrt {4{x}^{2}{y}^{4}·3xy}}$

$={\sqrt {4{x}^{2}{y}^{4}}·{\sqrt {3xy}}}$

$=2x{y}^{2}{\sqrt {3xy}}$

$解：{\sqrt {8{x}^{3}y}}$

$={\sqrt {4{x}^{2}·2xy}}$

$={\sqrt {4{x}^{2}}}·{\sqrt {2xy}}$

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

$解：原式 ={\sqrt {{x}^{4}(x+1)}}$

$={\sqrt {{x}^{4}}·}{\sqrt {x+1}}$

$={x}^{2}{\sqrt {x+1}}$

$解：原式 ={\sqrt {a({a}^{4}+2{a}^{2}{b}^{2}+{b}^{4})}}$

$={\sqrt {a{({a}^{2}+{b}^{2})}^{2}}}$

$={\sqrt {a}}·{\sqrt {{({a}^{2}+{b}^{2})}^{2}}}$

$=({a}^{2}+{b}^{2}){\sqrt {a}}$

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