Toán [toan 9] giá trị lớn nhất

D

dien0709

$t=\sqrt{x}\ge 0\to P=\dfrac{t}{t^2+t+1}$

$\to P=\dfrac{\dfrac{1}{3}(t^2+t+1)-\dfrac{1}{3}(t^2-2t+1)}{t^2+t+1}$

$P=\dfrac{1}{3}-\dfrac{(t-1)^2}{3[(t+\dfrac{1}{2})^2+\dfrac{3}{4}]}\le \dfrac{1}{3}$

dấu"="<=>t=1=>x=1
 
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