第106页 - 苏科版数学补充习题八年级上下册答案

√

×

√

$解：原式 =3{\sqrt {2}×2{\sqrt {6}}}$

$=(3×2)×({\sqrt {2}×{\sqrt {6}}})$

$=6×2{\sqrt {3}}$

$=12{\sqrt {3}}$

$解：原式 ={\sqrt {\frac {1} {2}×48}}$

$={\sqrt {24}}$

$=2{\sqrt {6}}$

$解：原式 ={\sqrt {105}÷({\sqrt {15}×{\sqrt {7}}})}$

$={\sqrt {105}}÷{\sqrt {15×7}}$

$={\sqrt {105}}÷{\sqrt {105}}$

$=1$

$解：原式 ={\sqrt {72÷12}}$

$={\sqrt {6}}$

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

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

$解：原式 ={\frac {\sqrt {1}} {\sqrt {3}}}$

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

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

$解：原式 ={\frac {\sqrt {1}} {\sqrt {b}}}$

$={\frac {\sqrt {1}×{\sqrt {b}}} {\sqrt {b}×{\sqrt {b}}}}$

$={\frac {\sqrt {b}} {b}}$

$解：原式 ={\frac {1×{\sqrt {a}}} {\sqrt {a}×{\sqrt {a}}}}$

$={\frac {\sqrt {a}} {a}}$