\(B=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)
Đề B lớn nhất thì \(x^2+1\) nhỏ nhất
Có: \(x^2+1\ge1\Rightarrow\dfrac{3}{x^2+1}\le\dfrac{3}{1}=3\)
Vậy \(MAX_B=3\) khi x = 0