Question

giúp mình với ;-;

Answer 1

a) \(A=\dfrac{3}{y+5}=\dfrac{3}{-2+5}=\dfrac{3}{-3}=-1\)

b) \(B=\dfrac{y\left(y-3\right)+2y\left(y+3\right)-3y^2-9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3y-9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3\left(y-3\right)}{\left(y-3\right)\left(y+3\right)}=\dfrac{3}{y+3}\)

c) \(M=-\dfrac{A}{B}=-\left(\dfrac{3}{y+5}:\dfrac{3}{y+3}\right)=-\dfrac{3\left(y+3\right)}{3\left(y+5\right)}=-\dfrac{y+3}{y+5}=-\dfrac{\left(y+5\right)-2}{y+5}=-1+\dfrac{2}{y+5}\in Z\)

\(\Rightarrow\left(y+5\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp ĐKXĐ:

\(\Rightarrow y\in\left\{-7;-6;-4\right\}\)