\(A=\dfrac{-3x^2+3}{4x^2+2}\)
\(A\in Z\Rightarrow-3x^2+3⋮4x^2+2\)
\(\Rightarrow-4\left(-3x^2+3\right)⋮4x^2+2\)
\(\Rightarrow12x^2-12⋮4x^2+2\)
\(\Rightarrow12x^2+6-18⋮4x^2+2\)
\(\Rightarrow3\left(4x^2+2\right)-18⋮4x^2+2\)
\(3\left(4x^2+2\right)⋮4x^2+2\Rightarrow18⋮4x^2+2\)
\(\Rightarrow4x^2+2\in U\left(18\right)\)
\(U\left(18\right)=\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)
Mà 4x2 +2 và luôn luôn lớn hơn hoặc=0 nên:
\(\left\{{}\begin{matrix}4x^2+2=2\Rightarrow4x^2=0\Rightarrow x^2=0\Rightarrow x=0\left(TM\right)\\4x^2+2=6\Rightarrow4x^2=4\Rightarrow x^2=1\Rightarrow x=\pm1\left(TM\right)\\4x^2+2=18\Rightarrow4x^2=16\Rightarrow x^2=4\Rightarrow x=\pm2\left(TM\right)\end{matrix}\right.\)
Vậy....
