Question

пожалуйста помогите решить уравнения​

Answer 1

Ответ:

$1)\ \dfrac{3a}{10}+\dfrac{2a}{10}=\dfrac{3a+2a}{10}=\dfrac{5a}{10}=\dfrac{a}{2};\\\\2)\ \dfrac{6x}{5y}-\dfrac{x}{5y}=\dfrac{6x-x}{5y}=\dfrac{5x}{5y}=\dfrac{x}{y};\\\\3)\ \dfrac{2m-4n}{21c}+\dfrac{5m+18n}{21c}=\dfrac{2m-4n+5m+18n}{21c}=\dfrac{7m-14n}{21c}=\dfrac{7(m-2n)}{21}=\dfrac{m-2n}{3};\\\\4)\ \dfrac{2a+5b}{ab}-\dfrac{2a-3b}{ab}=\dfrac{2a+5b-(2a-3b)}{ab}=\dfrac{2a+5b-2a+3b}{ab}=\dfrac{8b}{ab} =\dfrac{8}{a};$

$5)\ \dfrac{5y}{y^2-9} -\dfrac{15}{y^2-9} =\dfrac{5y-15}{y^2-9}=\dfrac{5(y-3)}{(y-3)(y+3)}=\dfrac{5}{y+3};\\\\6)\ \dfrac{y^2+8y}{4-y^2}-\dfrac{4y-4}{4-y^2}=\dfrac{y^2+8y-4y+4}{4-y^2}=\dfrac{y^2+4y+4}{4-y^2}=\dfrac{(y+2)^2}{(2-y)(2+y)}=\dfrac{(2+y)^2}{(2-y)(2+y)}=\dfrac{2+y}{2-y}.$