\(a.ĐKXĐ:x^2-16\ne0\Leftrightarrow x\ne\pm4\)
\(b.M=\frac{x^2-4x-3x+12}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{x\left(x-4\right)-3\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)
\(=\frac{\left(x-3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{x-3}{x-4}vớix\ne\pm4\)
\(c.\frac{x-3}{x-4}=1+\frac{1}{x-4}\)
Để \(x\in Z,M\in Z\)thì \(x-4\inƯ\left(1\right)\)
\(Ư\left(1\right)\in\left\{\pm1\right\}\)
\(TH1:x-4=1\)
\(x=5\)
\(TH2:x-4=-1\)
\(x=3\)