K
Khách

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

5 tháng 8 2016

nhầm = 16/45

5 tháng 8 2016

\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)

\(=\frac{4}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)

\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)

26 tháng 4 2017

\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)

\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)

\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)

\(=1\cdot3\cdot2\)

\(=6\)

\(A^2=6^2=36\)

21 tháng 10 2016

Ta có: \(\frac{1}{4\times6}=\frac{1}{4\times1\times3\times2}=\frac{1}{4\times3\times1\times2}\)

\(\frac{1}{8\times9}=\frac{1}{4\times2\times3\times3}=\frac{1}{4\times3\times2\times3}\)

\(\frac{1}{12\times12}=\frac{1}{4\times3\times3\times4}\)

\(\frac{1}{16\times15}=\frac{1}{4\times4\times3\times5}=\frac{1}{4\times3\times4\times5}\)......

\(\frac{1}{2680\times2013}=\frac{1}{4\times670\times3\times671}\)

Do đó:

\(M=\frac{1}{4\times3}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{670\times671}\right)\)

\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{670}-\frac{1}{671}\right)\)

\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{671}\right)=\frac{1}{12}\times\frac{670}{671}=\frac{335}{4026}\)

Vậy \(M=\frac{335}{4026}\)

23 tháng 4 2016

1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10

=1/2-1/10

=5/10-1/10

=4/10=2/5

23 tháng 4 2016

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}-\frac{1}{10}\)

\(\frac{2}{5}\)