\(\frac{x^2+2x+3}{x^2+2}=\frac{\frac{1}{2}\left(x^2+2\right)+\frac{1}{2}\left(x^2+4x+4\right)}{x^2+2}=\frac{1}{2}+\frac{\frac{1}{2}\left(x+2\right)^2}{x^2+2}\ge\frac{1}{2}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x=-2\)
\(\frac{x^2+2x+3}{x^2+2}=\frac{2\left(x^2+2\right)-\left(x^2-2x+1\right)}{x^2+2}=2-\frac{\left(x-1\right)^2}{x^2+2}\le2\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x=1\)
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