toán 9 tìm min

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transformers123

$A=\dfrac{8a^2+b}{4a}+b^2$

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

$\iff A=(\dfrac{a}{2}+\dfrac{b}{4a}+b^2)+\dfrac{3a}{2}$

$\iff A \ge 3\sqrt[3]{\dfrac{a}{2}.\dfrac{b}{4a}.b^2}+\dfrac{3a}{2}$

$\iff A \ge \dfrac{3b}{2}+\dfrac{3a}{2}$

$\iff A \ge \dfrac{3}{2}$

Vậy $\mathfrak{GTNN}$ của $A=\dfrac{3}{2}$ khi $x=y=\dfrac{1}{2}$
 
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