第103页 - 苏科版数学补充习题八年级上下册答案

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

$=-{\frac {7} {2}}{\sqrt {2}}$

$解：原式 =({\frac {\sqrt {2}} {4}-4{\sqrt {2}}})-({\frac {\sqrt {2}} {2}+{\sqrt {2}}})$

$={\frac {\sqrt {2}} {4}-{\frac {\sqrt {2}} {2}}}-4{\sqrt {2}-{\sqrt {2}}}$

$=-{\frac {\sqrt {2}} {4}}-5\sqrt {2}$

$=-{\frac {21} {4}}{\sqrt {2}}$

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

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

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

$解：原式 ={\sqrt {a}}-2a{\sqrt {b}}+4{\sqrt {a}}-{\sqrt {ab}}$

$=5{\sqrt {a}}-2a{\sqrt {b}}-{\sqrt {ab}}$

$解：原式 ={\frac {1} {\sqrt {2}}}×{\sqrt {6}}+{\sqrt {12}}×{\sqrt {6}}$

$={\sqrt {3}}+{\sqrt {72}}$

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

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

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