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Nam Lee
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Thiên Hàn
27 tháng 8 2018 lúc 19:32

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

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

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

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

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

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

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

Nguyễn Mạnh Hùng
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Nguyễn Đắc Định
8 tháng 4 2017 lúc 17:58

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

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

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

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

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

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

Hương Mai
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Phương Trâm
22 tháng 5 2017 lúc 20:58

Giải:

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

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

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

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

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

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

Phạm_Tiến_Đức
23 tháng 5 2017 lúc 6:24

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

\(=\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+....+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\div3\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\div3\)

\(=\dfrac{9}{20}\div3\)

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

Nguyễn Kim Thành
23 tháng 5 2017 lúc 9:44

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 hihi

Hương Mai
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Nguyễn Kim Thành
23 tháng 5 2017 lúc 9:34

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

Phương Trâm
22 tháng 5 2017 lúc 21:00

Mình giải cho bạn rồi mà.?

Từ Đào Cẩm Tiên
23 tháng 5 2017 lúc 6:07

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

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

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

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

= \(\dfrac{9}{60}=\dfrac{3}{20}\)

Nam Lee
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Nguyễn Lê Phước Thịnh
27 tháng 8 2022 lúc 19:53

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

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

 

Cỏ dại
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Núi non tình yêu thuần k...
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 Mashiro Shiina
9 tháng 12 2018 lúc 13:24

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

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

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

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

help me
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Akai Haruma
4 tháng 2 2023 lúc 14:23

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

 

Bùi Minh Anh
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nhok song tử
24 tháng 3 2017 lúc 18:22

tất nhên là bằng 00000000000000000000000000000000000000