K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 5 2015

(1+1/2)(1+1/3)...(1+1/n)

=3/2.4/3....n+1/n

=(3.4....n+1)/(2.3....n)=n+1/2

1/2+1/3+1/4+..+1/n<1/2.(n-1)=(n-1)/2   (do có n-1 số hạng) <n+1/2

=> lớn hơn

 

2 tháng 4 2023

1+1=3 :)))

14 tháng 4 2019

\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)

\(\Rightarrow T=\frac{1004}{1005}\)

14 tháng 4 2019

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\frac{2010}{2011}\)

\(\Rightarrow A=\frac{1005}{2011}\)

28 tháng 4 2016

D= [(1-1/2)(1-1/3)...(1-1/25)]:[(1+1/2)(1+1/3)...(1+1/25)]

D= [1/2. 2/3. ... . 24/25]: [3/2. 4/3. ... . 26/25]

D= 1/25 : 2/26

D= 1/25 . 26/2= 13/25

Vậy D= 13/25

28 tháng 4 2016

\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]\)\(:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)

\(D=\left[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{24}{25}\right]:\left[\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{26}{25}\right]\)

\(D=\frac{1.2.3...24}{2.3.4...25}:\frac{3.4.5...26}{2.3.4...25}\)

\(D=\frac{1}{25}:13\)

\(D=\frac{1}{325}\)

Tính giá trị biểu thức :1. \(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\) 2. \(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)3. \(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)4. \(D=(\frac {150}{1111}+\frac{5}{75}-\frac{14}{77})(\frac{1}{5}-\frac{1}{6}-\frac{1}{30})...
Đọc tiếp

Tính giá trị biểu thức :

1. \(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\) 

2. \(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)

3. \(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

4. \(D=(\frac {150}{1111}+\frac{5}{75}-\frac{14}{77})(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}) \)

5. Cho \(M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right);N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\). Tính \(P=M-N\)

6. \(E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)

7. \(F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

8. \(G=\left[\frac{\left(6-4\frac{1}{2}\right):0,03}{\left(3\frac{1}{20}-2,65\right).4+\frac{2}{5}}-\frac{\left(0,3-\frac{3}{20}\right).1\frac{1}{2}}{\left(1,88+2\frac{3}{25}\right).\frac{1}{80}}\right]:\frac{49}{60}\)

9. \(H=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)

10. \(I=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\)

11. \(k=\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{999}\right)\)

12. \(L=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}...\)(98 thừa số)

13. \(M=-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{3}}}}\)

14. \(N=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)

15. \(P=\left(\frac{1}{4}-1\right)\left(\frac{1}{5}-1\right)...\left(\frac{1}{2000}-1\right)\left(\frac{1}{2001}-1\right)\)

16. \(Q=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\right):\left(\frac{1}{1004.2006}+\frac{1}{1005.2005}+...+\frac{1}{2006.1004}\right)\)

3
2 tháng 5 2018

\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)

\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)

\(=\frac{2}{4}=\frac{1}{2}\)

\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)

\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)

\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)

\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)

\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)

\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)

\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)

\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\)               \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)

\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)                         \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)

\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\)                     \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)

\(=\frac{58}{7}-\frac{487}{63}\)                                          \(=\frac{577}{45}-\frac{280}{45}\)

\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\)                             \(=\frac{33}{5}\)

\(P=M-N\)

\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)

\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)

\(\Rightarrow P=\frac{-272}{45}\)

Vậy P = \(\frac{-272}{45}\)

\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)

\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)

\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)

\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)

\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)

\(=\frac{3}{8}+\frac{5}{8}=1\)

Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !

9 tháng 2 2018
sao không tự làm một số bài dễ đi
NV
12 tháng 5 2019

\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)

\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)

\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)

\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)

\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)

\(\Rightarrow-480\le x\le480\)

\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)

\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)

\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)

\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)

\(\Leftrightarrow x-3=\frac{90}{7}\)

\(\Rightarrow x=\frac{111}{7}\)

12 tháng 5 2019

???

25 tháng 3 2018

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)

\(\Rightarrow\)\(A< 1\) ( đpcm ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~