a: DKXĐ: x<>3; x<>-4
\(A=\dfrac{3x}{x+4}+\dfrac{2x}{3-x}+\dfrac{23x+8}{x^2+x-12}\)
\(=\dfrac{3x^2-9x-2x^2-8x+23x+8}{\left(x+4\right)\left(x-3\right)}=\dfrac{x^2+6x+8}{\left(x+4\right)\left(x-3\right)}\)
\(=\dfrac{x+2}{x+4}\)
b: x^2-16=0
=>x=-4(loại) hoặc x=4(nhận)
Khi x=4 thì \(A=\dfrac{4+2}{4+4}=\dfrac{6}{8}=\dfrac{3}{4}\)
c: Để A là số tự nhiên thì \(\left\{{}\begin{matrix}x+2⋮x+4\\\dfrac{x+2}{x+4}>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+4\in\left\{1;-1;2;-2\right\}\\\left[{}\begin{matrix}x>=-2\\x< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{-5;-2;-6\right\}\)
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