Tìm GTLN và GTNN

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vipboycodon

* $\dfrac{x^4+1}{(x^2+1)^2} = \dfrac{x^4+1}{x^4+2x^2+1}$

Ta có: $(x^2-1)^2 \ge 0 \rightarrow 2x^2 \le x^4+1$

$\rightarrow \dfrac{x^4+1}{x^4+2x^2+1} \ge \dfrac{1}{2}$

Vậy Min $= \dfrac{1}{2}$ khi $x^2 = 1 \leftrightarrow x = \pm 1$

* $\dfrac{x^4+1}{(x^2+1)^2} = \dfrac{-2x^2}{(x^2+1)^2}+1 \le 1$

Vậy Max $= 1$ khi $x = 0$
 
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