Đáp án:
$\begin{array}{l}
a)DKxd:{x^3} + {x^2} + x + 1 \ne 0\\
\Rightarrow {x^2}\left( {x + 1} \right) + \left( {x + 1} \right) \ne 0\\
\Rightarrow \left( {x + 1} \right)\left( {{x^2} + 1} \right) \ne 0\\
\Rightarrow x \ne - 1\left( {do:{x^2} + 1 > 0\forall x} \right)\\
b)Q = \frac{{3x + 3}}{{{x^3} + {x^2} + x + 1}}\\
= \frac{{3\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}\\
= \frac{3}{{{x^2} + 1}}\\
c)x = 0\left( {tmdk} \right)\\
\Rightarrow Q = \frac{3}{{{x^2} + 1}} = \frac{3}{1} = 3\\
d)Dkxd:x \ne - 1\\
Q = 3\\
\Rightarrow \frac{3}{{{x^2} + 1}} = 3\\
\Rightarrow {x^2} + 1 = 1\\
\Rightarrow x = 0\left( {tmdk} \right)\\
e)DKxd:x \ne - 1\\
Q = \frac{3}{{{x^2} + 1}} \in Z\\
\Rightarrow 3 \vdots {x^2} + 1\\
\Rightarrow Do:{x^2} + 1 \ge 1 \Rightarrow \left[ \begin{array}{l}
{x^2} + 1 = 3\\
{x^2} + 1 = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = 2\\
{x^2} = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \pm \sqrt 2 \left( {tm} \right)\\
x = 0\left( {tm} \right)
\end{array} \right.\\
Vậy\,x \in \left\{ { - \sqrt 2 ;0;\sqrt 2 } \right\}\\
f){x^2} + 1 \ge 1\forall x \ne - 1\\
\Rightarrow \frac{1}{{{x^2} + 1}} \le \frac{1}{1} = 1\forall x \ne - 1\\
\Rightarrow \frac{3}{{{x^2} + 1}} \le 3\forall x \ne - 1\\
\Rightarrow B \le 3\\
\Rightarrow GTLN:B = 3 \Leftrightarrow x = 0
\end{array}$
Vậy Giá trị lớn nhất của B=3 khi x=0
