Question

Алгебра, восьмой класс

Answer 1

Объяснение:

$(\frac{a+6}{3a+6} -\frac{1}{a+3} )*\frac{3}{a-3} -\frac{6}{a^2-9} =(\frac{a+6}{3*(a+3)} -\frac{1}{a+3} )*\frac{3}{a-3} -\frac{6}{a^2-9} =\frac{a+6-3}{3*(a+3)} *\frac{3}{a-3}-\frac{6}{a^2-9}=\\ =\frac{a+3}{3*(a+3)}*\frac{3}{a-3}-\frac{6}{a^2-9} =\frac{1}{a-3}-\frac{6}{(a+3)*(a-3)}=\frac{a+3-6}{(a+3)*(a-3)}=\frac{a-3}{(a+3)*(a-3)}= \\ =\frac{1}{a+3} =\frac{1}{-\frac{1}{4}+3 }=\frac{1}{2\frac{3}{4} }=\frac{1}{\frac{11}{4} }=\frac{4}{11} .$