Chứng minh : \(\dfrac{9}{22}\) < \(\dfrac{1}{4}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...\(\dfrac{1}{100}\)<\(\dfrac{9}{10}\)
Cho \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}.\) Chứng minh rằng: \(S>\dfrac{9}{22}\)
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
\(\dfrac{-22}{121}\) = \(\dfrac{4}{-22}\) = - \(\dfrac{10}{55}\)
Chứng minh các phân số sau bằng nhau.
\(\dfrac{-22}{121}=\dfrac{-22:11}{121:11}=\dfrac{-2}{11}\)
\(\dfrac{4}{-22}=\dfrac{4:2}{-22:2}=\dfrac{2}{-11}=\dfrac{-2}{11}\)
\(-\dfrac{10}{55}=-\dfrac{10:5}{55:5}=-\dfrac{2}{11}=\dfrac{-2}{11}\)
Vậy ...
Khi rút gọn tất cả bằng -2 phần 11 nên các phân số trên bằng nhau
a , cho A = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\) . Chứng minh A < \(\dfrac{7}{4}\)
b ,cho B = 21 + 22 + 23 + ... + 260 . Chứng minh B \(⋮\) 21
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
Chứng minh: \(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+\(\dfrac{1}{24}\)+....+\(\dfrac{1}{80}\)không phải là một số tự nhiên.
1)
Cho \(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{200}\)
Chứng minh: \(A>\dfrac{9}{10}\)
2)
Cho \(B=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
Chứng minh: \(B>\dfrac{7}{12}\)
HELP ME!!!!!!!!
\(B=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...\dfrac{1}{200}\right)>\dfrac{1}{150}+..\dfrac{1}{150}+\dfrac{1}{200}+..+200=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)Vậy ... (ta có điều phải chứng minh )
Ta có :\(\dfrac{1}{20}>\dfrac{1}{200}\)
...
\(\dfrac{1}{199}>\dfrac{1}{200}\)
Do đó : \(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+..+\dfrac{1}{200}=\dfrac{181}{200}>\dfrac{180}{200}=\dfrac{9}{10}\)Vậy ...
So sánh :
a) Chứng minh rằng : M = \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.......+\dfrac{1}{100!} \)
Chứng minh rằng : M <1 .
b) Chứng minh rằng : N = \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+........+\dfrac{9}{1000!}\)
Chứng minh rằng : N < \(\dfrac{1}{9!}\)
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
Chứng minh rằng: \(\dfrac{1 + 3 + 5 + ... + 39}{21 + 22 + 23 + ... + 40} = \dfrac{1}{2^{20}}\)
Trong các phân số \(\dfrac{5}{4}\);\(\dfrac{22}{23}\);\(\dfrac{9}{9}\);\(\dfrac{24}{23}\) phân số bé nhất là
A.\(\dfrac{24}{23}\) B.\(\dfrac{22}{23}\) C.\(\dfrac{5}{4}\) D.\(\dfrac{9}{9}\)
`=>` `B`
Vì các phân số cồn lại thuộc dạng `(x >= 1)`
Chứng minh rằng: \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+...+\dfrac{9}{1000!}< \dfrac{1}{9!}\)
Ta có:
\(\dfrac{9}{n!}\)< \(\dfrac{n-1}{n!}\) = \(\dfrac{1}{(n-1)!} - \dfrac{1}{n!}\) với n > 10 (n thuộc Z)
\(\Rightarrow\) \(\dfrac{9}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!} \)
= \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!}\)
\(\Rightarrow\) \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{1}{10!} - \dfrac{1}{11!} + \dfrac{1}{11!} - \dfrac{1}{12!} + ....\)
= \(\dfrac{1}{9!} - \dfrac{1}{1000!}\)
\(\Rightarrow \) \(\dfrac{9}{10!} + \dfrac{9}{11!} + ...+ \dfrac{9}{1000!} < \dfrac{1}{9!}\)
Chúc bn hc tốt.