a: ĐKXĐ: x|x+2|-x^2+4<>0
TH1: x>=-2
BPT sẽ là x^2+2x-x^2+4<>0
=>x<>-2
=>x>-2
TH2: x<-2
BPT sẽ là -x^2-2x-x^2+4<>0
=>-2x^2-2x+4<>0
=>x^2+x-2<>0
=>x<>-2; x<>1
=>x<-2
b: \(D=\dfrac{x\left(x+2\right)\left(x-1\right)}{x\left|x+2\right|-x^2+4}\)
TH1: x>-2
\(D=\dfrac{x\left(x+2\right)\left(x-1\right)}{x^2+2x-x^2+4}=\dfrac{x\left(x-1\right)}{2}\) luôn là só nguyên
TH2: x<-2
\(D=\dfrac{x\left(x+2\right)\left(x-1\right)}{-2x^2-2x+4}=\dfrac{-x\left(x+2\right)\left(x-1\right)}{2\left(x^2+x-2\right)}\)
\(=\dfrac{-x\left(x+2\right)\left(x-1\right)}{2\left(x+2\right)\left(x-1\right)}=\dfrac{-x}{2}\)
Để D nguyên thì x=2k
c: Khi x=6 thì \(D=\dfrac{6^3+6^2-2\cdot6}{6\left|6+2\right|-6^2+4}=15\)
