\(A=\dfrac{1}{\left|x-3\right|+\left|x^2-4\right|}\)
ĐKXD: \(\left|x-3\right|+\left|x^2-4\right|>0\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne\pm2\end{matrix}\right.\)
\(B=\dfrac{1}{\left|x-2\right|+\left|4-x\right|-2}\)
ĐKXD: \(\left|x-2\right|+\left|4-x\right|\ne2\)
Ta có: \(\left|x-2\right|+\left|4-x\right|\ge\left|x-2+4-x\right|=2\)
Như vậy,nếu không xảy ra \(2\le x\le4\) thì thỏa mãn