Tìm min
$H=\sqrt{x^{2}+1}+\sqrt{x^{2}-2x+5}$
Edited by kimchitwinkle, 21-09-2016 - 20:22.
Tìm giá trị nhỏ nhất $H=\sqrt{x^{2}+1}+\sqrt{x^{2}-2x+5}$
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Started By lelehieu2002, 19-09-2016 - 21:02
#3
Posted 20-09-2016 - 12:48
Iceghost
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$H = \sqrt{x^2 + 1} + \sqrt{(1-x)^2 + 2^2} \overset{Mincowski}{\geqslant} \sqrt{(x + 1 -x)^2 + (1 + 2)^2} = \sqrt{10}$
$\implies H_\text{min} = \sqrt{10} \iff x= \dfrac13$
Edited by Iceghost, 20-09-2016 - 12:51.
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