\(C=\dfrac{5-x^2}{x^2+3}=\dfrac{-x^2-3+8}{x^2+3}=-1+\dfrac{8}{x^2+3}\)
Ta có: \(x^2>=0\forall x\)
=>\(x^2+3>=3\forall x\)
=>\(\dfrac{8}{x^2+3}< =\dfrac{8}{3}\forall x\)
=>\(\dfrac{8}{x^2+3}-1< =\dfrac{8}{3}-1=\dfrac{5}{3}\forall x\)
=>\(C< =\dfrac{5}{3}\forall x\)
Dấu '=' xảy ra khi x2=0
=>x=0
Vậy: \(C_{Max}=\dfrac{5}{3}\) khi x=0