\(ĐK:x\ne-1\)
\(E=\dfrac{4x-5}{x+1}=\dfrac{4\left(x+1\right)-9}{x+1}=4-\dfrac{9}{x+1}\)
Để `E` nguyên thì \(\dfrac{9}{x+1}\in Z\)\(\Rightarrow x+1\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
`@x+1=1->x=0(tm)`
`@x+1=-1->x=-2(tm)`
`@x+1=3->x=2(tm)`
`@x+1=-3->x=-4(tm)`
`@x+1=9->x=8(tm)`
`@x+1=-9->x=-10(tm)`
Vậy \(x\in\left\{0;-2;2;-4;8;10\right\}\) thì `E` nguyên