Question

tìm min max cuar C = \(\dfrac{x^2+2x+3}{x^2+2}\)

Answer 1

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Tìm min:

\(C=\dfrac{x^2+2x+3}{x^2+2}\\ =\dfrac{\dfrac{1}{2}\left(x^2+2\right)+\dfrac{x^2}{2}+2x+2}{x^2+2}\\ =\dfrac{1}{2}+\dfrac{2\left(\dfrac{x}{2}+1\right)^2}{x^2+2}\\ Vì\dfrac{2\left(\dfrac{x}{2}+1\right)^2}{x^2+2}\ge0\forall x\\ \Rightarrow C\ge\dfrac{1}{2}\\ \Rightarrow Min_C=\dfrac{1}{2}\Leftrightarrow x=-2\)

Tìm Max:

\(C=\dfrac{x^2+2x+3}{x^2+2}\\ =\dfrac{2\left(x^2+2\right)-x^2+2x-1}{x^2+2}\\ =2-\dfrac{\left(x-1\right)^2}{x^2+2}\\ Vì\dfrac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\\ \Rightarrow C\le2\\ \Rightarrow Max_C=2\Leftrightarrow x=1\)