Ta có : \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}=\frac{1}{2}-\frac{2}{27}=\frac{23}{54}\)

Trả lời:

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{1}{2}-\frac{2}{27}\)

\(=\frac{23}{54}\)

Học tốt 

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)

B=1+\(\dfrac{1}{5.10}\)+\(\dfrac{1}{10.15}\)+\(\dfrac{1}{15.20}\)+......+\(\dfrac{1}{95.100}\)

5B = 5 +\(\dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+........+\dfrac{5}{95.100}\)

5B=5+\(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+.........+\dfrac{1}{95}-\dfrac{1}{100}\)

5B=5+\(\dfrac{1}{5}-\dfrac{1}{100}\)

5B=\(\dfrac{519}{100}\)

=>B= \(\dfrac{519}{100}:5=\dfrac{519}{500}\)

A= \(\dfrac{1}{3.7}\) +\(\dfrac{1}{7.11}\)+\(\dfrac{1}{11.15}\)+\(\dfrac{1}{15.19}\)+\(\dfrac{1}{19.23}\)+\(\dfrac{1}{23.27}\)

A= 4.(\(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+\dfrac{1}{15.19}+\dfrac{1}{19.23}\)+\(\dfrac{1}{23.27}\)

A=4.\(\dfrac{1}{3.7}+4.\dfrac{1}{7.11}+4.\dfrac{1}{11.15}+4.\dfrac{1}{15.19}+4.\dfrac{1}{19.23}+4.\dfrac{1}{23.27}\)

A=\(4.(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{29}+\dfrac{1}{29})\)

A= 4 (.\(\dfrac{1}{3}-\dfrac{1}{29}\))

A=\(\dfrac{104}{87}\)

a) Mình ko ghi lại đề nhé!

= \(\frac{1}{2}\) - ( \(\frac{1}{3.7}\) + \(\frac{1}{7.11}\) + ... + \(\frac{1}{23.27}\) )

= \(\frac{1}{2}\) - \(\frac{1}{4}\) . ( \(\frac{1}{3}\) -  \(\frac{1}{7}\) + \(\frac{1}{7}\) - .... - \(\frac{1}{27}\) )

= \(\frac{1}{2}\) - \(\frac{1}{4}\) . ( \(\frac{1}{3}\) - \(\frac{1}{27}\) )

= \(\frac{1}{2}\) - \(\frac{1}{4}\) . \(\frac{8}{27}\)

= \(\frac{1}{2}\) - \(\frac{2}{27}\) = \(\frac{23}{54}\)

b) ..............................................................................

= \(\frac{1}{5}\)  . ( \(\frac{5}{5.10}\) - \(\frac{5}{10.15}\) - ... - \(\frac{5}{95.100}\) )

= \(\frac{1}{5}\) . ( \(\frac{1}{5}\) - \(\frac{1}{10}\) + \(\frac{1}{10}\) - ... - \(\frac{1}{100}\) )

= \(\frac{1}{5}\) . ( \(\frac{1}{5}\) - \(\frac{1}{100}\) )

= \(\frac{1}{5}\) . \(\frac{19}{100}\)

= \(\frac{19}{500}\) 

k mình nha! Chúc bạn học tốt và được nhiều k!

\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)

= \(4.\left(\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)

=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)

= \(1.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)

=\(1.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)

=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)

= \(1.\dfrac{8}{27}\)

= \(\dfrac{8}{27}\)

bạn tham khảo nha, cách làm như vậy đó

Câu hỏi của Nguyễn Thị Mai Ca - Toán lớp 7 - Học toán với OnlineMath 

ban kia lam dung roi do

k tui nha 

thanks

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)

\(=\frac{8}{27}\)

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+.....+\frac{27-23}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)