第112页 - 苏科版数学补充习题八年级上下册答案

$ \frac{3\sqrt{2}}{2}$

$ 30\sqrt{2}$

3

$21-4\sqrt{5}$

$2{\sqrt {10}}+4{\sqrt {5}}$

$ -3a$

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

$ 8或-8$

$解：原式 =({\frac {2} {3}×{\frac {3} {4}}})×({\sqrt {6}×{\sqrt {24}}})$

$\ \ \ \ \ \ \ \ ={\frac {1} {2}}×12$

$\ \ \ \ \ \ \ \ =6$

$解：原式 =(12{\sqrt {3}-6{\sqrt {3}}})÷{\sqrt {3}}$

$\ \ \ \ \ \ \ \ =6{\sqrt {3}}÷{\sqrt {3}}$

$\ \ \ \ \ \ \ \ =6$

$解：原式 ={\sqrt {10}}×3{\sqrt {\frac {2} {5}}}-{\sqrt {10}×{\sqrt {\frac {5} {2}}}}$

$ =3×{\sqrt {10×{\frac {2} {5}}}}-{\sqrt {10×{\frac {5} {2}}}}$

$ =3×2-5$

$ =1$

$解：原式 =6{\sqrt {3}}÷({\sqrt {3}-2{\sqrt {3}}})$

$\ \ \ \ \ \ \ \ =6{\sqrt {3}}÷(-{\sqrt {3}})$

$\ \ \ \ \ \ \ \ =-6$