Question

Tìm GTLN:

\(\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}\)

Answer 1

\(L=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+1\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\dfrac{3}{x^2+1}\le3\)

Dấu "=" xảy ra khi: \(x=0\)

Answer 2

\(A=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)

do \(x^2\ge0\forall x\)

=>\(x^2+1\ge1\)

=>\(\dfrac{3}{x^2+1}\le3\)

=>A≤3

Max A=3 dấu = xảy ra khi

x=0