\(C=\dfrac{10}{7\cdot12}+\dfrac{10}{12\cdot17}+...+\dfrac{10}{502\cdot507}\)

\(=2\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+...+\dfrac{1}{502}-\dfrac{1}{507}\right)\)

\(=2\cdot\dfrac{500}{3549}=\dfrac{1000}{3549}\)

a,A=1/5-1/8+1/8-1/11+...+1/2006-1/2009=1/5-1/2009=2004/10045

b,B=1/4x(4/6x10+4/10x14+...+4/402x406)

=1/4x(1/6-1/10+1/10-1/14+...+1/402-1/406)

=1/4x(1/6-1/406)

=1/4x100/609=25/609

c,C=2x(5/7x12+5/12x17+...+5/502x507)

=2x(1/7-1/12+1/12-1/17+...+1/502-1/507)

=2x(1/7-1/507)

=2x500/3549

=1000/3549

Xin lỗi vì ko viết được rõ ràng.Mong bạn thông cảm. Chúc bạn học tốt.

\(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{2006\times2009}\)

\(=\frac{1}{3}\left(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{2006\times2009}\right)\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\right)\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{2009}\right)\)

\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{2009}\right)\)

\(=\frac{1}{3}\left(\frac{2009}{10045}-\frac{5}{10045}\right)\)

\(=\frac{1}{3}.\frac{2004}{10045}=\frac{2004}{30135}\)

a)

\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)

\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)

b)

\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)

\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)

\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)

\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)

\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)

c)

\(C=\frac{10}{7,12}+\frac{10}{12.17}+...+\frac{10}{502.507}\)

\(\Rightarrow \frac{C}{2}=\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\)

\(\frac{C}{2}=\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{507-502}{502.507}\)

\(\frac{C}{2}=\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{502}-\frac{1}{507}\)

\(\frac{C}{2}=\frac{1}{7}-\frac{1}{507}=\frac{500}{3549}\)

\(\Rightarrow C=\frac{1000}{3549}\)

Giúp với 

Ta có:\(\dfrac{20}{2\times7}+\dfrac{20}{7\times12}+\dfrac{20}{12\times17}+\dfrac{20}{17\times22}+\dfrac{20}{22\times27}+\dfrac{20}{27\times32}\)

 \(=4\times\left(\dfrac{5}{2\times7}+\dfrac{5}{7\times12}+\dfrac{5}{12\times17}+\dfrac{5}{17\times22}+\dfrac{5}{22\times27}+\dfrac{5}{27\times32}\right)\)

    \(=4\times\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{32}\right)\)

\(=4\times\left(\dfrac{1}{2}-\dfrac{1}{32}\right)=4\times\dfrac{15}{32}=\dfrac{30}{16}\)

\(\dfrac{20}{2\cdot7}+\dfrac{20}{7\cdot12}+\dfrac{20}{12\cdot17}+\dfrac{20}{17\cdot22}+\dfrac{20}{22\cdot27}+\dfrac{20}{27\cdot32}\) ( * là nhân nha )

20 x ( \(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}+\dfrac{1}{22\cdot27}+\dfrac{1}{27\cdot32}\) )

20 x ( \(\dfrac{1}{2}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{12}-\dfrac{1}{12}+\dfrac{1}{17}-\dfrac{1}{17}+\dfrac{1}{22}-\dfrac{1}{22}+\dfrac{1}{27}-\dfrac{1}{27}+\dfrac{1}{32}\) )

20 x ( \(\dfrac{1}{2}-\dfrac{1}{32}\) ) 

20 x \(\dfrac{15}{32}\) = \(\dfrac{300}{32}=\dfrac{75}{8}\)

\(A=\dfrac{10}{3\cdot7}-\dfrac{1}{7}+\dfrac{1}{12}-\dfrac{1}{12}+\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{1}{24}=\dfrac{10}{21}+\dfrac{1}{24}=\dfrac{29}{56}\)

bn nên gõ latex để mn dễ hỗ trợ trl nha

\(\dfrac{3}{8}\) 

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trả lời cẩn thận đi minato

thế các bạn có trả lời đàng hoàng giùm mình ko

dua thoi gap gap lam mình can so sánh kẹt qua

\(\frac{6}{7.12}+\frac{6}{12.17}+...+\frac{6}{87.92}+\frac{6}{92.95}\)

= \(6\left(\frac{5}{7.12}.\frac{1}{5}+\frac{5}{12.17}.\frac{1}{5}+...+\frac{5}{92.95}.\frac{1}{5}\right)\)

= \(6.\frac{1}{5}\left(\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{87.92}+\frac{5}{92.95}\right)\)

= \(\frac{6}{5}\left(\frac{5}{7}-\frac{5}{12}+\frac{5}{12}-\frac{5}{17}+...+\frac{5}{92}-\frac{5}{95}\right)\)

= \(\frac{6}{5}\left(\frac{5}{7}-\frac{5}{95}\right)\)= \(\frac{6}{5}.\frac{88}{133}=\frac{528}{665}\)

Tự rút gọn, mình lười.

Chọn đáp án D

Em cần phần nào nhỉ .

A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)

A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)

A = \(\dfrac{105}{106}\)

B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)

B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)

C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))

C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)

C = \(\dfrac{5}{51}\) 

D = \(\dfrac{1}{2}\) +   \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)

D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)

D = \(\dfrac{8}{9}\)

E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))

E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)

E = \(\dfrac{147}{200}\)