a: \(D=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
\(ĐK:x\ne0;x\ne-5\\ a,D=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\\ D=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}=\dfrac{x-1}{2}\\ b,D=\dfrac{1}{4}\Leftrightarrow2\left(x-1\right)=1\Leftrightarrow x=\dfrac{3}{2}\left(tm\right)\\ c,D< 0\Leftrightarrow x-1< 0\left(2>0\right)\Leftrightarrow x< 1;x\ne0\\ d,\dfrac{1}{D}\in Z^-\Leftrightarrow\dfrac{2}{x-1}\in Z^-\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-1;0\right\}\left(\dfrac{1}{D}< 0\right)\)
