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\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}+\frac{1}{10.10}\)
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.10}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
\(A>1+0+0+0+...+0-\frac{1}{10}\)
\(A>1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow A>\frac{5}{10}=\frac{1}{2}\)
mà : \(\frac{1}{2}=\frac{66}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\)
Ta có: \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Rightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
- Đến đây bn lấy \(\frac{9}{22}\) so sánh vs \(\frac{65}{132}\) là ra ĐPCM nhé :3
Câu 1:
\(S=\frac{10}{7}+\frac{10}{7^2}+\frac{10}{7^3}+...+\frac{10}{7^{10}}\)
\(\frac{1}{7}S=\frac{10}{7^2}+\frac{10}{7^3}+....+\frac{10}{7^{11}}\)
\(\rightarrow\)\(\left(1-\frac{1}{7}\right).S=\frac{10}{7}-\frac{10}{7^{11}}\)
=> \(S=\frac{10.7^{10}-10}{7^{10}.6}\)
Ta có:
1/2 + 1/3 + 1/4 + ... + 1/15 + 1/16 = (1/2 + 1/3 + 1/4 + 1/5) + (1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11) + (1/12 + 1/13 + 1/14) + (1/15 + 1/16)
Vì 1/6 + 1/7 + 1/8 < 3x 1/6 = 1/2
1/9 + 1/10 + 1/11 <3x1/9 = 1/3
1/12 + 1/13 +1/14 < 3x1/12 = 1/4
1/15 + 1/16 < 3 x 1/15 = 1/5
Nên A < 2 x (1/2 + 1/3 + 1/4 + 1/5) < 2 x (1/2 + 1/2 + 1/4 + 1/4) =3 (1)
Lập luận tương tự có:
A = ( 1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12) + (1/13 + 1/14 + 1/15 + 1/16) > (1/2 + 1/3 + 1/4) + 4 x 1/8 + 4 x 1/ 12 + 4 x 1/16
Hay A > 2 x (1/2 + 1/3 + 1/4) > 2 x (1/2 + 1/4 + 1/4) = 2 (2)
Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.
1)
\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)
= \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)
= \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)
= \(\frac{2^8.\left(-6\right)}{2^8}\)
= \(-6\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2023\cdot2024}\)
\(=\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{2023}-\frac{1}{2024}\)
\(=\frac11-\frac{1}{2024}=\frac{2023}{2024}\)
A=1⋅21+2⋅31+3⋅41+⋯+2023⋅20241
\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \hdots + \frac{1}{2023} - \frac{1}{2024}\)
\(= \frac{1}{1} - \frac{1}{2024} = \frac{2023}{2024}\)
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