a: \(D=\dfrac{3x^2+6x+2x-4-14x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x\left(x-1\right)}\)
\(=\dfrac{3x^2-6x}{\left(x-2\right)}\cdot\dfrac{1}{x\left(x-1\right)}=\dfrac{3}{x-1}\)
b: x-1-3=0
=>x=4(nhận)
KHi x=4 thì \(A=\dfrac{3}{4-1}=\dfrac{3}{3}=1\)
c: Để Dlà số nguyên thì \(x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{0;4\right\}\)