CMR \(\dfrac{1}{4}\)<\(\dfrac{2-\sqrt{2+\sqrt{2+\sqrt{2+..............+\sqrt{2}}}}}{2-\sqrt{2+\sqrt{2+..............+\sqrt{2}}}}\)<\(\dfrac{3}{10}\)
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DL
1
2
8 tháng 5 2022
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)
Ta có ;
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2012.2013}\)
\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(\Leftrightarrow B< 1-\dfrac{1}{2013}\)
\(\Rightarrow B< \dfrac{2012}{2013}\)
Lại có : \(\dfrac{2012}{2013}< \dfrac{3}{4}\)
\(\Rightarrow B< \dfrac{3}{4}\)
* Chắc vậy, sai thì thôg cảm ^^ *
Còn j k hiểu thì ib nha
27 tháng 6 2021
a, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )
\(\Rightarrow a^2+2ab+b^2=\left(a+b\right)^2\ge4ab\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
b, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )
\(\Rightarrow ab\le\dfrac{a^2+b^2}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 5 2023
Hôm nay olm sẽ hướng dẫn các em mẹo giải các bài toán dạng này như sau:
Ta thấy vế phải là \(\dfrac{1}{2}\) thì vế trái sẽ ≤ \(\dfrac{1}{2}\) - a ( a > 0)
Em biến đổi mẫu số các phân số lần lượt thành lũy thừa của các số tự nhiên liên tiếp. Sau đó rút gọn tổng các phân số đó thì sẽ chứng minh được em nhé.
A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\)
A = \(\dfrac{1}{\left(1.2\right)^2}\)+\(\dfrac{1}{\left(2.2\right)^2}\)+\(\dfrac{1}{\left(2.3\right)^2}\)+...+\(\dfrac{1}{\left(2.50\right)^2}\)
A = \(\dfrac{1}{1^2.2^2}\)+\(\dfrac{1}{2^2.2^2}\)+\(\dfrac{1}{2^2.3^2}\)+...+\(\dfrac{1}{2^2.50^2}\)
A = \(\dfrac{1}{2^2}\)\(\times\)(\(\dfrac{1}{1^2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{50^2}\))
A = \(\dfrac{1}{4}\) \(\times\)(1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+...+\(\dfrac{1}{50.50}\))
Vì \(\dfrac{1}{1}\)> \(\dfrac{1}{2}\)>\(\dfrac{1}{3}\)>\(\dfrac{1}{4}\)>...>\(\dfrac{1}{50}\)
⇒ \(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{50.50}\)<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...\(\dfrac{1}{49.50}\)
A = \(\dfrac{1}{4}\).(1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+\(\dfrac{1}{4.4}\)+..+\(\dfrac{1}{50.50}\)) < \(\dfrac{1}{4}\) .(1+\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+..+\(\dfrac{1}{49.50}\))
A < \(\dfrac{1}{4}\).(1+\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\))
A<\(\dfrac{1}{4}\).(2 - \(\dfrac{1}{50}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{200}\) < \(\dfrac{1}{2}\)
Vậy A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\) < \(\dfrac{1}{2}\) ( đpcm)
AH
Akai Haruma
Giáo viên
25 tháng 12 2018
Lời giải:
Ta có:
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{(2n)^2}< \frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}(*)\)
Mà:
\(\frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n-1)(2n+1)}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+1}\right)\)
\(< \frac{1}{6}< \frac{1}{4}(**)\)
Từ \((*);(**)\Rightarrow N< \frac{1}{4}\) (đpcm)
18 tháng 9 2021
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{3^2}< \dfrac{1}{2\cdot3};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \RightarrowĐpcm\)
5 tháng 11 2021
\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)
\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)
\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)