Tìm GTNN. GTLN

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iceghost

Bài 3
$\dfrac{2015}{x^2-x+1}=\dfrac{2015}{(x-\dfrac12)^2+\dfrac34}$
$Ta \; có \; : \; (x-\dfrac12)^2+\dfrac34 \ge \dfrac34 \\
\implies \dfrac{2015}{(x-\dfrac12)^2+\dfrac34} \le \dfrac{2015}{\dfrac34}=\dfrac{8060}3$
Vậy $Max = \dfrac{8060}3 \iff x-\dfrac12=0 \iff x=\dfrac12$
 
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