Ta có :

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

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}==\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

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

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

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

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

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+\frac{4}{15\cdot19}+\frac{4}{19\cdot23}+\frac{4}{23.27}\)

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

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

\(=\frac{4}{4}\cdot\frac{8}{27}=\frac{8}{27}\)

Gọi số cần tìm là \(x\), ta có :

S = \(\frac{4}{3x7}\)+  \(\frac{4}{7x11}\)+ \(\frac{4}{11x15}\)+ ............\(x\) = \(\frac{664}{1995}\)

 = \(\frac{4}{3}\)- \(\frac{4}{7}\)+ \(\frac{4}{7}\) -  \(\frac{4}{11}\)+  \(\frac{4}{11}\) -  \(\frac{4}{15}\)+ ..............\(x\) = \(\frac{664}{1995}\)

 = \(\frac{4}{3}\)-  \(x\)=  \(\frac{664}{1995}\)( loại các sô giống nhau )

\(x\)= \(\frac{4}{3}\)-  \(\frac{664}{1995}\)

\(x\)=  \(\frac{1996}{1995}\)

a.Goi so cuoi la x ta co

....................(de bai)

=1/3-1/7+1/7-1/11+1/11-1/15+...-x=664/1995

=1/3-x=664/1995

x=1/3-664/1995

x=1/1995

BN HOANG HIEU OI BN TINH HOI NHAM RUI!BN THU XEM LAI HIEU CUA MAU SO SO VOI TU SO DI!!!

\(A=\frac{4}{3X7}+\frac{4}{7X11}+\frac{4}{11X15}+...+\frac{4}{100X104}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(=\frac{1}{3}-\frac{1}{104}\)

\(=\frac{101}{312}\)

Chúc bạn học giỏi nha!!!

K cho mik với nhé nguyen huu thuong 2005

\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(A=\frac{1}{3}-\frac{1}{104}=\frac{104}{312}-\frac{3}{312}=\frac{101}{312}\)

A=4/3.7 +4/7.11 +4/11.15 +...+4/100.104 

A=1/3 −1/7 +1/7 −1/11 +1/11 −1/15 +...+1/100 −1/104 

A=1/3 −1/104 =104/312 −3/312 =101/312 

\(C=\frac{4}{3x7}+\frac{4}{7x11}+\frac{4}{11x15}+\frac{4}{15x19}+\frac{4}{19x23}+\frac{4}{23x27}\)

     = 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27

     =1/3-(1/7+1/7)-(1/11+1/11)-(1/15-1/15)-(1/19+1/19)-(1/23-1/23)-1/27

    =1/3-1/27

   =...

=8/27

Ta có: \(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{23\cdot27}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{23}-\dfrac{1}{27}\)

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

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(=\frac{1}{3}-\frac{1}{104}=\frac{104}{312}-\frac{3}{312}=\frac{101}{312}\)

\(A=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\)(Dấu . là dấu nhân)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}\)
\(=\dfrac{9}{27}-\dfrac{1}{27}\)
\(=\dfrac{8}{27}\)

A = 4/3x7 + 4/7x11+ 4/11x15 + 4/15x19 + 4/19 x23 + 4/23 x 27

A = 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23 -1/27

A = 1/3 - 1/27

A = 8/27

lớp 4 mà học dạng này rồi sao!?

\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)

=\(3\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{12}{43\cdot47}\right)=3\left(\frac{1}{3}\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{11}-\frac{1}{11}\right)-...-\frac{1}{47}\right)\)

=\(3\left(\frac{1}{3}-\frac{1}{47}\right)=3\cdot\frac{44}{141}=\frac{44}{47}\)

Nếu đúng cho xin lik nhé