Bài 2:
\(a,B=\dfrac{-5\left(x-3\right)+18}{x-3}=-5+\dfrac{18}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(18\right)=\left\{-2;-1;1;2;3;6;9;18\right\}\left(x\in N\right)\\ \Leftrightarrow x\in\left\{1;2;4;5;6;9;12;21\right\}\\ b,x_{min}\Leftrightarrow\left(x-3\right)_{min}\Leftrightarrow x-3=-18\Leftrightarrow x=-15\)
Bài 3:
\(a,M=\dfrac{x^2+3+x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x}=\dfrac{x\left(x+1\right)}{x\left(x+3\right)}=\dfrac{x+1}{x+3}=1-\dfrac{2}{x+3}\in N\\ \Leftrightarrow x+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-5;-4;-2;-1\right\}\)
Thay vào ta thấy \(x\in\left\{-5;-4;-1\right\}\Leftrightarrow M\in N\)
\(b,M_{max}\Leftrightarrow1-\dfrac{2}{x+3}\) đạt max \(\Leftrightarrow\dfrac{2}{x+3}\) đạt min \(\Leftrightarrow x+3\) đạt max
Mà \(x\in Z\Leftrightarrow x=-1\Leftrightarrow M_{max}=1-\dfrac{2}{-1+3}=0\)
\(c,M=??\)
