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23 tháng 5 2017

Đặt A= \(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

A= \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\) 3A = \(3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\) 3A= \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\) 3A= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\) 3A= \(\dfrac{1}{2}-\dfrac{1}{20}\)

3A= \(\dfrac{9}{20}\)

A= \(\dfrac{9}{20}:3\)

A = \(\dfrac{3}{20}\)

Vậy A= \(\dfrac{3}{20}\)

Chúc bạn học tốt haha

22 tháng 5 2017

Mình giải cho bạn rồi mà.?

27 tháng 8 2018

\(C=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)

\(C=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\right)\)

\(C=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(C=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)

\(C=\dfrac{1}{3}.\dfrac{9}{20}\)

\(C=\dfrac{3}{20}\)

23 tháng 8 2018

kết quả cuối cùng là : 3/20

23 tháng 8 2018

Gọi \(S=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)

Nhân hai vế với 3 và áp dụng công thức tách một phân số thành hiệu hai phân số:

\(\dfrac{x}{n\left(n+x\right)}=\dfrac{1}{n}-\dfrac{1}{n+x}\)

\(\Rightarrow3S=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\)

\(\Rightarrow3S=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\)

\(\Rightarrow3S=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\)

\(\Rightarrow3S=\dfrac{1}{2}-\dfrac{1}{20}\)

\(\Rightarrow3S=\dfrac{10}{20}-\dfrac{1}{20}\)

\(\Rightarrow3S=\dfrac{9}{20}\)

\(\Rightarrow S=\dfrac{9}{20}:3\)

\(\Rightarrow S=\dfrac{9}{20}.\dfrac{1}{3}\)

\(\Rightarrow S=\dfrac{3}{20}\)

8 tháng 4 2017

Đặt \(A=\dfrac{1}{10}-\dfrac{1}{40}-\dfrac{1}{88}-\dfrac{1}{154}-\dfrac{1}{238}-\dfrac{1}{340}\)

\(\Leftrightarrow3A=\dfrac{3}{10}-\dfrac{3}{40}-\dfrac{3}{88}-\dfrac{3}{154}-\dfrac{3}{238}-\dfrac{3}{340}\)

\(\Leftrightarrow3A=\dfrac{3}{2\cdot5}-\dfrac{3}{5\cdot8}-\dfrac{3}{8\cdot11}-\dfrac{3}{11\cdot14}-\dfrac{3}{14\cdot17}-\dfrac{3}{17\cdot20}\)

\(\Leftrightarrow3A=\left(\dfrac{1}{2}-\dfrac{1}{5}\right)-\left(\dfrac{1}{5}-\dfrac{1}{8}\right)-\left(\dfrac{1}{8}-\dfrac{1}{11}\right)-\left(\dfrac{1}{11}-\dfrac{1}{14}\right)-\left(\dfrac{1}{14}-\dfrac{1}{17}\right)-\left(\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{8}-\dfrac{1}{8}+\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{14}-\dfrac{1}{14}+\dfrac{1}{17}-\dfrac{1}{17}+\dfrac{1}{20}\)

\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{20}\\ \Leftrightarrow3A=\dfrac{3}{20}\\ \Leftrightarrow A=\dfrac{1}{20}\)

22 tháng 5 2017

Giải:

Đặt \(A=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)

\(A=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(A=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)

\(A=\dfrac{1}{3}.\dfrac{9}{20}\)

\(A=\dfrac{3}{20}\)

22 tháng 5 2017

\(\dfrac{1}{10}\)+\(\dfrac{1}{40}\)+\(\dfrac{1}{88}\)+\(\dfrac{1}{154}\)+\(\dfrac{1}{238}\)+\(\dfrac{1}{340}\)

=\(\dfrac{1}{2.5}\)+\(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+\(\dfrac{1}{14.17}\)+\(\dfrac{1}{17.20}\)

=\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{11}\)-\(\dfrac{1}{14}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{14}\)-\(\dfrac{1}{17}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{17}\)-\(\dfrac{1}{20}\))

=\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{14}\)+\(\dfrac{1}{14}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)+\(\dfrac{1}{20}\))

=\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\)+\(\dfrac{1}{20}\))

=\(\dfrac{1}{3}\).\(\dfrac{10+1}{20}\)

=\(\dfrac{1}{3}\).\(\dfrac{11}{20}\)

=\(\dfrac{11}{60}\)

a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)

b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)

c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)

 

9 tháng 12 2018

\(\dfrac{1}{x^2+7x+10}+\dfrac{1}{x^2+13x+40}+\dfrac{1}{x^2+19x+88}+\dfrac{1}{x^2+25x+154}\)

\(=\dfrac{1}{\left(x+2\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+8\right)}+\dfrac{1}{\left(x+8\right)\left(x+11\right)}+\dfrac{1}{\left(x+11\right)\left(x+14\right)}\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+11}+\dfrac{1}{x+11}-\dfrac{1}{x+14}\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}\)

9 tháng 12 2018

sai rồi !!! xem lại đi nhé =))

24 tháng 3 2017

tất nhên là bằng 00000000000000000000000000000000000000

AH
Akai Haruma
Giáo viên
4 tháng 2 2023

Lời giải:
Gọi tử số là $T$

\(T=(1-\frac{1}{6})+(1-\frac{2}{7})+(1-\frac{3}{8})+....+(1-\frac{88}{93})\)

\(=\frac{5}{6}+\frac{5}{7}+\frac{5}{8}+....+\frac{5}{93}=5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})\)

Gọi mẫu số là $M$
\(M=\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+....+\frac{1}{93})\)

Do đó:
\(C=\frac{5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}{\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}=\frac{5}{\frac{-1}{2}}=-10\)