so sánh $\sqrt[]{2007}$ + $\sqrt[]{2009}$ với 2.$\sqrt[]{2008}$ 14/08/2021 Bởi Adeline so sánh $\sqrt[]{2007}$ + $\sqrt[]{2009}$ với 2.$\sqrt[]{2008}$
Ta có: $+) \, (\sqrt{2007} + \sqrt{2009})^2$ $= 2007 + 2009 + 2\sqrt{2007.2009}$ $= 2008 -1 + 2008 – 1 + 2\sqrt{(2008-1)(2008+1)}$ $= 2.2008 + 2\sqrt{2008^2 -1}$ $+) \, (2\sqrt{2008})^2$ $= 4.2008$ $=2.2008 + 2.2008$ $= 2.2008 + 2.\sqrt{2008^2}$ Do $2008^2 > 2008^2 – 1$ $\Rightarrow \sqrt{2008^2} > \sqrt{2008^2 -1}$ $\Rightarrow 2.\sqrt{2008^2} > 2\sqrt{2008^2 -1}$ $\Rightarrow 2.2008 + 2.\sqrt{2008^2} > 2.2008 + 2\sqrt{2008^2 -1}$ $\Rightarrow (2\sqrt{2008})^2 > (\sqrt{2007} + \sqrt{2009})^2$ $\Rightarrow 2\sqrt{2008} > \sqrt{2007} + \sqrt{2009}$ Bình luận
Ta có:
$+) \, (\sqrt{2007} + \sqrt{2009})^2$
$= 2007 + 2009 + 2\sqrt{2007.2009}$
$= 2008 -1 + 2008 – 1 + 2\sqrt{(2008-1)(2008+1)}$
$= 2.2008 + 2\sqrt{2008^2 -1}$
$+) \, (2\sqrt{2008})^2$
$= 4.2008$
$=2.2008 + 2.2008$
$= 2.2008 + 2.\sqrt{2008^2}$
Do $2008^2 > 2008^2 – 1$
$\Rightarrow \sqrt{2008^2} > \sqrt{2008^2 -1}$
$\Rightarrow 2.\sqrt{2008^2} > 2\sqrt{2008^2 -1}$
$\Rightarrow 2.2008 + 2.\sqrt{2008^2} > 2.2008 + 2\sqrt{2008^2 -1}$
$\Rightarrow (2\sqrt{2008})^2 > (\sqrt{2007} + \sqrt{2009})^2$
$\Rightarrow 2\sqrt{2008} > \sqrt{2007} + \sqrt{2009}$