Question

8класс. 99 . следующие выражения: 1. 2. 3. 4.

Answer 1

$\boxed{1}\ \ \frac{b+3}{b^3+9b}\cdot (\frac{b+3}{b-3} + \frac{b-3}{b+3})= \frac{b+3}{b^3+9b}\cdot( \frac{b^2+6b+9+b^2-6b+9}{b^2-9} )=\\\\ = \frac{b+3}{b(b^2+9)}\cdot\frac{2b^2+18}{b^2-9} =\frac{b+3}{b(b^2+9)}\cdot \frac{2(b^2+9)}{(b-3)(b+3)}= \frac{2}{b(b-3)}$

$\boxed{2}\ \ \frac{x^2-9}{2x^2+1}\cdot (\frac{6x+1}{x-3}+ \frac{6x-1}{x+3})= \frac{x^2-9}{2x^2+1}\cdot( \frac{(6x+1)(x+3)+(6x-1)(x-3)}{x^2-9} )=\\\\ =\frac{x^2-9}{2x^2+1}\cdot \frac{6(2x^2+1)}{x^2-9}=6$

$\boxed{3}\ \ (\frac{st}{s^2-t^2}+ \frac{t}{2t-2s})\cdot \frac{s+t}{2t}= (\frac{st}{(s-t)(s+t)}- \frac{t}{2(s-t)})\cdot \frac{s+t}{2t}= \frac{2st-t(s+t)}{2(s^2-t^2)} \cdot \frac{s+t}{2t}=\\\\ = \frac{t(s-t)}{2(s-t)(s+t)}\cdot \frac{s+t}{2t}= \frac{1}{4}$

$\boxed{4}\ \ \frac{3a+b}{a^2b-ab^2}+ \frac{b-a}{ab}: \frac{a^2-b^2}{3a-b}= \frac{3a+b}{ab(a-b)}- \frac{a-b}{ab}\cdot \frac{3a-b}{(a-b)(a+b)}=\\\\ =\frac{3a+b}{ab(a-b)}- \frac{3a-b}{ab(a+b)}= \frac{(3a+b)(a+b)-(3a-b)(a-b)}{ab(a-b)(a+b)}= \frac{8ab}{ab(a-b)(a+b)}= \frac{8}{a^2-b^2}$