Question

Помогите решить пожалуйста 6 7.

Answer 1

$1)(\frac{b}{a}- \frac{a}{b})* \frac{1}{b+a} = \frac{b^{2}-a^{2}}{ab}* \frac{1}{b+a}= \frac{(b-a)(b+a)}{ab} *\frac{1}{a+b}= \frac{b-a}{ab}\\\\\frac{\frac{1}{3}-1} {1*\frac{1}{3}}=- \frac{2}{3}: \frac{1}{3}=- \frac{2}{3}*3=-2$

$2)(\frac{1}{9a}+ \frac{1}{3a} )*\frac{a^{2}} {8} =\frac{1+3}{9a}* \frac{a^{2}} {8}= \frac{4*a^{2}} {9a*8}= \frac{a}{18}\\\\\frac{9}{18}= \frac{1}{2}=0,5$