Căn bậc hai số học

F

fcb_provip@yahoo.com

P

pe_lun_hp

2a.

$(\sqrt{2014} - \sqrt{2013})(\sqrt{2014} + \sqrt{2013}) = 1$

$\Rightarrow \sqrt{2014} - \sqrt{2013} = \dfrac{1}{\sqrt{2014} + \sqrt{2013}}$

Tương tự:

$\sqrt{2013} - \sqrt{2012} = \dfrac{1}{\sqrt{2013} + \sqrt{2012}}$

Hiển nhiên :

$\sqrt{2014} + \sqrt{2013} > \sqrt{2013} + \sqrt{2012}$

Nên :

$\dfrac{1}{\sqrt{2014} + \sqrt{2013}} < \dfrac{1}{\sqrt{2013} + \sqrt{2012}}$

Hay:

$\sqrt{2014} - \sqrt{2013} < \sqrt{2013} - \sqrt{2012}$
 
P

pe_lun_hp

2b.

Áp dụng hằng đẳng thức :

$\sqrt{a + \sqrt{b}} - \sqrt{a - \sqrt{b}} = \sqrt{2(a - \sqrt{a^2 - b})}$

$\sqrt{3+\sqrt{5}} - \sqrt{3-\sqrt{5}} - \sqrt{2} = \sqrt{2(3 - \sqrt{3^2 - 5})} - \sqrt{2} = 0$

Vậy : $\sqrt{3+\sqrt{5}} - \sqrt{3-\sqrt{5}} - \sqrt{2} = 0$
 
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