\(\dfrac{x^2-3}{x^2-1}=\dfrac{x^2-1-2}{x^2-1}=\dfrac{x^2-1}{x^2-1}-\dfrac{2}{x^2-1}=\\ 1-\dfrac{2}{x^2-1}\)để \(\dfrac{x^2-3}{x^2-1}\in Z\Leftrightarrow x^2-1\inƯ\left(2\right)=\left(-1;1;-2;2\right)\)
\(x^2-1\) | -1 | 1 | -2 | 2 |
x | 0 | \(\varnothing\) | \(\varnothing\) | \(\varnothing\) |
vậy x = 0 thì biểu thức \(\dfrac{x^2-3}{x^2-1}\in Z\)