Cho $a>-1; b>-1; c>-4; a+b+c =0$. Tìm max của $A =\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{4+c}$
Edited by perfectstrong, 25-01-2024 - 19:43.
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Tìm max A = $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{4+c}$
Started By lekarlsen, 25-01-2024 - 18:36
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Posted 25-01-2024 - 18:36
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*bang bang* maxwell's silver hammer came down upon her head
*clang clang* maxwell's silver hammer made sure she was dead
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Posted 25-01-2024 - 19:56
hxthanh
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Tín đồ $\sum$
$A=3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{4}{4+c}\right)\le 3-\left(\dfrac{4}{2+a+b}+\dfrac{4}{4+c}\right)$Cho $a>-1; b>-1; c>-4; a+b+c =0$. Tìm max của $A =\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{4+c}$
$\le 3-\dfrac{16}{6+a+b+c}=\dfrac 13$
Dấu bằng xảy ra khi $a=b=\frac 12, c=-1$
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