Tìm Min,Max

T

transformers123

$\bigstar$ Tìm max:

Ta có:

$2a^2+\dfrac{b^2}{4}+\dfrac{1}{a^2}=4$

$\iff a^2+a^2+\dfrac{b^2}{4}+\dfrac{1}{a^2}=4$

$\iff 4\sqrt[4]{\dfrac{a^4b^2}{4a^2}} \le 4$

$\iff 4\sqrt[4]{\dfrac{a^2b^2}{4}} \le 4$

$\iff \sqrt{\dfrac{ab}{2} } \le 1$

$\iff \dfrac{ab}{2} \le 1$

$\iff ab \le 2$

$\iff ab+2014 \le 2016$

Dấu "=" xảy ra khi $a=1,\ b=2$
 
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