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\(\dfrac{4}{3.5}+\dfrac{8}{5\cdot9}+\dfrac{12}{9\cdot15}+...+\dfrac{32}{n\left(n+6\right)}=\dfrac{16}{25}\)

\(2\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+2\left(\dfrac{1}{5}-\dfrac{1}{9}\right)+2\left(\dfrac{1}{9}-\dfrac{1}{15}\right)+...+2\left(\dfrac{1}{n}-\dfrac{1}{n+16}\right)=\dfrac{16}{25}\)

\(2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{n}-\dfrac{1}{n+16}\right)=\dfrac{16}{25}\)

\(2\left(\dfrac{1}{3}-\dfrac{1}{n+16}\right)=\dfrac{16}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{1}{n+16}=\dfrac{1}{75}\)

⇒ \(n+16=75\) 

\(\Rightarrow n=59\)

20 tháng 4 2017

\(\frac{4}{3.5}+\frac{8}{5.9}+\frac{12}{9.15}+...+\frac{32}{n\left(n+16\right)}=\frac{16}{25}\)

\(2\left(\frac{1}{3}-\frac{1}{5}\right)+2\left(\frac{1}{5}-\frac{1}{9}\right)+2\left(\frac{1}{9}-\frac{1}{15}\right)+...+2\left(\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)

\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)

\(2\left(\frac{1}{3}-\frac{1}{n+16}\right)=\frac{16}{25}\)

\(\frac{1}{3}-\frac{1}{n+16}=\frac{8}{25}\)

\(\frac{1}{n+16}=\frac{1}{75}\)

\(\Rightarrow n+16=75\)

\(\Rightarrow n=59\)

28 tháng 3

bạn  Thanh Tùng DZ mình vẫn chưa hiểu tại sao \(\dfrac{8}{5.9}\) = 2.(\(\dfrac{1}{5}\) - \(\dfrac{1}{9}\)

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
25 tháng 7 2016

m) (\(\frac{-5}{12}\)+\(\frac{6}{11}\))+(\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\))

\(\frac{-5}{12}\)+\(\frac{6}{11}\)+\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\)

= (\(\frac{-5}{12}\)+\(\frac{5}{12}\))+(\(\frac{6}{11}\)+\(\frac{5}{11}\))+\(\frac{7}{17}\)

= 0+1+\(\frac{7}{17}\)

\(\frac{24}{17}\)

n) (\(\frac{9}{16}\)+\(\frac{8}{-27}\))+(1+\(\frac{7}{16}\)+\(\frac{-19}{27}\))

\(\frac{9}{16}\)+\(\frac{8}{-27}\)+1+\(\frac{7}{16}\)+\(\frac{-19}{27}\)

= (\(\frac{9}{16}\)+\(\frac{7}{16}\))+(\(\frac{8}{-27}\)+\(\frac{-19}{27}\))+1

= 1+(-1)+1

= 0+1

= 1

o) (6-2\(\frac{4}{5}\)).3\(\frac{1}{8}\)-1\(\frac{3}{5}\):\(\frac{1}{4}\)

= (6-\(\frac{14}{5}\)).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)

\(\frac{16}{5}\).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)

= 10-\(\frac{8}{5}\):\(\frac{1}{4}\)

= 10-\(\frac{32}{5}\)

\(\frac{18}{5}\)

CHÚC BẠN HỌC TỐT

 

tung từng vế một thôi

bạn nhác quá éo chịu suy nghĩ

bài này dễ vl

13 tháng 5 2017

Bài 1:

a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)

\(\frac{1}{5x+6}=\frac{1}{2011}\)

=> 5x + 6 = 2011

    5x = 2011 - 6

    5x = 2005

    x = 2005 : 5

    x = 401

b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)

\(\frac{7}{x}=\frac{7}{15}\)

=> x = 15

c, ghi lại đề

d, ghi lại đề

Bài 2:

\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)

21 tháng 10 2018

ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac

28 tháng 1 2019

\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)

\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)

\(A=\frac{4-1}{36+12}\)

\(A=\frac{3}{48}=\frac{1}{16}\)

2 tháng 8 2017

Đề bị sai 

2 tháng 8 2017

Sửa đề . \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{71}{216}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)=\frac{71}{216}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{n+2}\right)=\frac{71}{216}\)

\(\Leftrightarrow\frac{1}{n+2}=1-\frac{71}{216}\div\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{n+2}=\frac{37}{108}\)

\(\Leftrightarrow x=\frac{34}{37}\Rightarrow\text{(đề sai) }\)

13 tháng 2 2018

\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)

\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)

\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)

\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=3-\left(1-\frac{1}{8}\right)\)

\(A=3-\frac{5}{8}\)

\(A=\frac{19}{8}\)