Q = \(\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)để Q đạt GTLN => \(x^2+1phảiNhỏnhất\)
\(x^2+1\ge1=>x^2+1\)đạt GTNN là 1 khi x=0
vậy Q đạt GTLN =3 khi x = 0
\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}\\ Q=\dfrac{3}{x^2+1}\\ Do\text{ }x^2\ge0\forall x\\ \Rightarrow x^2+1\ge1\forall x\\ Q=\dfrac{3}{x^2+1}\le3\forall x\\ \text{Dấu “=” xảy ra khi : }\\ x^2=0\\ \Leftrightarrow x=0\\Vậy\text{ }Q_{\left(Max\right)}=3\text{ }khi\text{ }x=0\)
