a, Ta có: \(A=\left(\frac{x+4}{3x+6}-\frac{1}{x^2+4x+4}\right).\left(1+\frac{x-1}{x+5}\right)\)
\(=\left(\frac{x+4}{3\left(x+2\right)}-\frac{1}{\left(x+2\right)^2}\right).\left(\frac{x+5}{x+5}+\frac{x-1}{x+5}\right)\)
\(=\left(\frac{\left(x+4\right)\left(x+2\right)}{3\left(x+2\right)^2}-\frac{1.3}{3\left(x+2\right)^2}\right).\frac{x+5+x-1}{x+5}\)
\(=\frac{\left(x+4\right)\left(x+2\right)-3}{3\left(x+2\right)^2}.\frac{2x+4}{x+5}\)
\(=\frac{x^2+2x+4x+8-3}{3\left(x+2\right)^2}.\frac{2\left(x+2\right)}{x+5}\)
\(=\frac{x^2+6x+5}{3\left(x+2\right)^2}.\frac{2\left(x+2\right)}{x+5}\)
\(=\frac{\left(x+1\right)\left(x+5\right)}{3\left(x+2\right)^2}.\frac{2\left(x+2\right)}{x+5}\)
\(=\frac{\left(x+1\right)\left(x+5\right).2\left(x+2\right)}{3\left(x+2\right)^2\left(x+5\right)}\) \(=\frac{2\left(x+1\right)}{3\left(x+2\right)}\)
b, Với \(x\ne-2,x\ne-5\) ta có:
\(A=\frac{2\left(x+1\right)}{3\left(x+2\right)}=\frac{2}{3}.\frac{x+1}{x+2}=\frac{2}{3}.\frac{\left(x+2\right)-1}{x+2}=\frac{2}{3}.1.\frac{-1}{x+2}=\frac{2}{3}.\frac{-1}{x+2}\)
Để A có giá trị là một số nguyên \(\Leftrightarrow\frac{2}{3}.\frac{-1}{x+2}\) có giá trị là một số nguyên \(\Leftrightarrow\frac{-1}{x+2}\in Z\) (vì \(\frac{2}{3}\in Z\))\(\Leftrightarrow-1⋮\left(x+2\right)\)
\(\Leftrightarrow x+2\inƯ\left(-1\right)=\left\{1;-1\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3\right\}\)
Đối chiếu ĐKXĐ \(\Rightarrow x\in\left\{-1;-3\right\}\)
Vậy để A có giá trị là một số nguyên thì \(x\in\left\{-1;-3\right\}\)
