第104页 - 苏科版数学补充习题八年级上下册答案

$解：原式 ={\sqrt {5}×{\sqrt {2}}-{\sqrt {3}}×{\sqrt {2}+{\sqrt {2}}}}×{\sqrt {2}}$

$={\sqrt {10}-{\sqrt {6}}+2}$

解：原式$ ={\sqrt {12}×{\sqrt {12}}}-$

${\sqrt {3}}×{\sqrt {12}}+1×{\sqrt {12}}$

$ =12-6+2{\sqrt {3}}$

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

$解：原式 =2{\sqrt {3}}×{\sqrt {3}}-{\sqrt {5}×{\sqrt {3}}}+$

$2{\sqrt {3}}×{\sqrt {5}}-{\sqrt {5}}×{\sqrt {5}}$

$=6-{\sqrt {15}}+2{\sqrt {15}}-5$

$=1+{\sqrt {15}}$

$解：原式 ={\frac {\sqrt {2}} {2}}×{\frac {1} {\sqrt {2}}}-{\frac {\sqrt {3}} {2}}×{\frac {1} {\sqrt {2}}}+$

${\frac {\sqrt {2}} {2}}×{\frac {1} {\sqrt {3}}}-{\frac {\sqrt {3}} {2}×{\frac {1} {\sqrt {3}}}}$

$={\frac {1} {2}}-{\frac {\sqrt {6}} {4}}+{\frac {\sqrt {6}} {6}}-{\frac {1} {2}}$

$=-{\frac {\sqrt {6}} {12}}$

$解：原式 ={\sqrt {3}}×5+{\sqrt {2}}×5-$

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

$=5{\sqrt {3}}+5{\sqrt {2}}-6{\sqrt {2}}-4{\sqrt {3}}$

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

$解：原式 =7×{\sqrt {5}}+{\sqrt {10}}×{\sqrt {5}}-$

$7×{\sqrt {2}}-{\sqrt {10}}×{\sqrt {2}}$

$=7{\sqrt {5}}+5{\sqrt {2}}-7{\sqrt {2}}-2{\sqrt {5}}$

$=5{\sqrt {5}}-2{\sqrt {2}}$

$解：原式 ={({\sqrt {5}})}^{2}-{({\sqrt {2}})}^{2}$

$=5-2$

$=3$

$解：原式 ={(2{\sqrt {3}})}^{2}-{({\sqrt {7}})}^{2}$

$=12-7$

$=5$

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

$2×{\sqrt {a}}×{\sqrt {b}}$

$ =a+b-2{\sqrt {ab}}$

$解：原式 ={({\sqrt {3}})}^{2}+{({\sqrt {2}})}^{2}+2×{\sqrt {3}}×{\sqrt {2}}$

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

$=5+2{\sqrt {6}}$