$\frac{2\sqrt{\frac{x^{2}}{2}+\frac{x}{2}+1}-x}{\sqrt{x-1}-x}\geq 1$
$\frac{2\sqrt{\frac{x^{2}}{2}+\frac{x}{2}+1}-x}{\sqrt{x-1}-x}\geq 1$
Started By Mr0, 15-04-2012 - 19:01
#1
Posted 15-04-2012 - 19:01
#2
Posted 15-04-2012 - 19:10
Điều kiện $x\geq1$
nhận xét $\sqrt{x-1}<x$
Chuyển 1 sang thì có
$\frac{\sqrt{2x^2+2x+4}-\sqrt{x-1}}{\sqrt{x-1}-x}\geq0$
tương đương
$\sqrt{2x^2+2x+4}\leq\sqrt{x-1}$ (do mẫu âm)
......
nhận xét $\sqrt{x-1}<x$
Chuyển 1 sang thì có
$\frac{\sqrt{2x^2+2x+4}-\sqrt{x-1}}{\sqrt{x-1}-x}\geq0$
tương đương
$\sqrt{2x^2+2x+4}\leq\sqrt{x-1}$ (do mẫu âm)
......
Edited by Giang1994, 15-04-2012 - 19:24.
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