Để phân số \(\dfrac{x+14}{x+11}\in Z\) thì :
\(x+14⋮x+11\)
Mà \(x+11⋮x+11\)
\(\Leftrightarrow3⋮x+11\)
\(\Leftrightarrow x+11\inƯ\left(3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+11=1\\x+11=3\\x+11=-1\\x+11=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-8\\x=-12\\x=-14\end{matrix}\right.\)
Vậy ..
A=\(\dfrac{x+14}{x+11}=\dfrac{x+11+3}{x+11}=1+\dfrac{3}{x+11}\)
để A nguyên <=> x+11 thuộc Ư (3)={ 1,-1,3,-3}
xét bảng giá trị
x+11 | 1 | -1 | 3 | -3 |
x | -10 | -12 | -8 | -14 |
vậy x ∈{-10;-12;-8;-14}