a: ĐKXĐ: x<>-m
=>TXĐ: D=R\{-m}
\(y=\dfrac{mx-2m+15}{x+m}\)
=>\(y'=\dfrac{\left(mx-2m+15\right)'\left(x+m\right)-\left(mx-2m+15\right)\left(x+m\right)'}{\left(x+m\right)^2}\)
\(=\dfrac{m\left(x+m\right)-mx+2m-15}{\left(x+m\right)^2}\)
\(=\dfrac{m^2+2m-15}{\left(x+m\right)^2}\)
Để hàm số đồng biến trên từng khoảng xác định là \(y'>0\forall x\in TXĐ\)
=>\(\dfrac{m^2+2m-15}{\left(x+m\right)^2}>0\)
=>\(m^2+2m-15>0\)
=>(m+5)(m-3)>0
TH1: \(\left\{{}\begin{matrix}m+5>0\\m-3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>3\\m>-5\end{matrix}\right.\)
=>m>3
TH2: \(\left\{{}\begin{matrix}m+5< 0\\m-3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -5\\m< 3\end{matrix}\right.\)
=>m<-5
b: TXĐ: D=R\{-m}
\(y=\dfrac{mx+4m}{x+m}\)
=>\(y'=\dfrac{\left(mx+4m\right)'\left(x+m\right)-\left(mx+4m\right)\left(x+m\right)'}{\left(x+m\right)^2}\)
\(=\dfrac{m\left(x+m\right)-mx-4m}{\left(x+m\right)^2}\)
\(=\dfrac{mx+m^2-mx-4m}{\left(x+m\right)^2}=\dfrac{m^2-4m}{\left(x+m\right)^2}\)
Để hàm số đồng biến trên từng khoảng xác định thì \(y'>0\forall x\)
=>\(\dfrac{m^2-4m}{\left(x+m\right)^2}>0\)
=>\(m^2-4m>0\)
=>\(m\left(m-4\right)>0\)
TH1: \(\left\{{}\begin{matrix}m>0\\m-4>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>0\\m>4\end{matrix}\right.\)
=>m>4
TH2: \(\left\{{}\begin{matrix}m< 0\\m-4< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 0\\m< 4\end{matrix}\right.\)
=>m<0
