Question

Нечётные решите пожалуйста

Answer 1

$19)\; \; \dfrac{9b}{a-b}\cdot \dfrac{a^2-ab}{45b}=\dfrac{a}{5}=\dfrac{-83}{5}=-16,6\\\\21)\; \; \dfrac{y(x+y)}{18y}\cdot \dfrac{9x}{x+y}=\dfrac{x}{2}=\dfrac{-9,6}{2}=-4,8\\\\23)\; \; \dfrac{a+5x}{a}:\dfrac{x(a+5x)}{a^2}=\dfrac{a+5x}{a}\cdot \dfrac{a^2}{x(a+5x)}=\dfrac{a}{x}=\dfrac{-74}{-10}=7,4$

$25)\; \; \dfrac{c\, (6-c)}{1-c}:\dfrac{c^2}{1-c}=\dfrac{c\, (6-c)}{1-c}\cdot \dfrac{1-c}{c^2}=\dfrac{6-c}{c}=\dfrac{6-1,2}{1,2}=4\\\\27)\; \; \Big(\dfrac{1}{5a}+\dfrac{1}{7a}\Big)\cdot \dfrac{a^2}{4}=\dfrac{12a}{35a^2}\cdot \dfrac{a^2}{4}=\dfrac{3a}{35}=\dfrac{3\cdot 7,7}{35}=0,66\\\\29)\; \; (x-6):\dfrac{x^2-12x+36}{x+6}=\dfrac{(x-6)\cdot (x+6)}{(x-6)^2}=\dfrac{x+6}{x-6}=\dfrac{-10+6}{-10-6}=0,25$

$31)\; \; \dfrac{a^2-36b^2}{6ab}:\Big(\dfrac{1}{6b}-\dfrac{1}{a}\Big)=\dfrac{(a-6b)(a+6b)}{6ab}:\dfrac{a-6b}{6ab}=\\\\=\dfrac{(a-6b)(a+6b)}{6ab}\cdot \dfrac{6ab}{a-6b}=a+6b=5\dfrac{5}{17}+6\cdot 5\dfrac{2}{17}=\dfrac{90}{17}+\dfrac{522}{17}=36\\\\33)\; \; \dfrac{7ab}{a+7b}\cdot \Big(\dfrac{a}{7b}-\dfrac{7b}{a}\Big)=\dfrac{7ab}{a+7b}\cdot \dfrac{a^2-(7b)^2}{7ab}=\dfrac{7ab}{a+7b}\cdot \dfrac{(a-7b)(a+7b)}{7ab}=\\\\=a-7b=(7\sqrt2+7)-7\cdot (\sqrt2-9)=7\sqrt2+7-7\sqrt2+63=70$