Chứng minh
1. [tex]\sqrt{2000}-2\sqrt{2001}+\sqrt{2002}[/tex] < 0
2. [tex]\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{(n+1)\sqrt{n}}[/tex] < 2
Bài 1 :
$(\sqrt{2000} + \sqrt{2002})^2 \\
= 2000 + 2002 + 2\sqrt{2000.2002} \\
= 4002 + 2\sqrt{2001^2 - 1} < 4002 + 2\sqrt{2001^2} = 2.2001 + 2.2001 = 4 . 2001 = (2\sqrt{2001})^2$
Suy ra : $(\sqrt{2000} + \sqrt{2002})^2 < (2\sqrt{2001})^2$
Hay $\sqrt{2000} + \sqrt{2002} < 2\sqrt{2001}$
Suy ra : $\sqrt{2000} + \sqrt{2002} - 2\sqrt{2001} < 0$
Bài 2 :
$\dfrac{1}{(n+1).\sqrt{n}} \\
=\dfrac{\sqrt{n}}{(n+1).n}\\
= \sqrt{n} . \dfrac{1}{(n+1).n} \\
= \sqrt{n}(\dfrac{1}{n} - \dfrac{1}{n + 1}) \\
= \sqrt{n}(\dfrac{1}{\sqrt{n}} + \dfrac{1}{\sqrt{n + 1}})(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n + 1}}) \\
= (1 + \dfrac{\sqrt{n}}{\sqrt{n + 1}})(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n + 1}}) < 2(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n + 1}})$
Áp dụng bđt trên vào bài ta sẽ c/m được
$\dfrac{1}{2\sqrt{1}} +\dfrac{1}{3\sqrt{2}} +\dfrac{1}{4\sqrt{3}}+....+\dfrac{1}{(n+1).\sqrt{n}} <2$
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