Min $A=\frac{x^2}{x+1}$ khi $x \geq 2$
Min $A=\frac{x^2}{x+1}$ khi $x \geq 2$
Started By bovuotdaiduong, 07-04-2016 - 16:20
#1
Posted 07-04-2016 - 16:20
bovuotdaiduong
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#2
Posted 07-04-2016 - 16:33
leminhnghiatt
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Min $A=\frac{x^2}{x+1}$ khi $x \geq 2$
$A=\dfrac{x^2}{x+1}+\dfrac{4(x+1)}{9}-\dfrac{4(x+1)}{9} \geq \dfrac{4x}{3}-\dfrac{4x}{9}-\dfrac{4}{9}=\dfrac{8x}{9}-\dfrac{4}{9} \geq \dfrac{16}{9}-\dfrac{4}{9}=\dfrac{4}{3}$
$MinA=\dfrac{4}{3} \iff x=2$
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