a) \(A=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)
b) Để A nhận gt nguyên thì 3 \(⋮x^2+1\)
\(\Rightarrow x^2+1\inƯ\left(3\right)\)
\(\Rightarrow x^2+1\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow x^2\in\left\{0;2;-2;-4\right\}\)
Vì \(x^2\ge0\Rightarrow\left[{}\begin{matrix}x^2=0\\x^2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
c) Ta có : \(x^2+1\ge1\) với mọi x
\(\Rightarrow\dfrac{3}{x^2+1}\le\dfrac{3}{1}=3\)
Dấu '=' xảy ra \(\Leftrightarrow x^2=0\\\Leftrightarrow x=0\)
Vậy Amax=3 \(\Leftrightarrow x=0\)
