M=1/3.7+1/7.11+1/11.15+...+1/43.47

M=1/3-1/7+1/7-1/11+1/11-1/15+...+1/43-1/47

M=1/3-1/47

CÒN LẠI TỰ TÍNH NHA BN

AI THẤY ĐÚNG THÌ ỦNG HỘ NHA

M=\(\frac{1}{3x7}+\frac{1}{7x11}+\frac{1}{11x15}+...+\frac{1}{43x47}\)

=>4M=\(\frac{4}{3x7}+\frac{4}{7x11}+...+\frac{4}{43x47}\)

=>4M=\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\)

=>4M=\(\frac{1}{3}-\frac{1}{47}\)

=>4M=\(\frac{44}{141}\)

=>M=\(\frac{44}{141}:4\)

=>M=\(\frac{11}{141}\)

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

Sao ko ai giup vây HuHu

?????

=12/3+12/11+12/15+..+12/43+12/47

=4+12/11+3/5+..+12/43+12/47

\(A=\frac{2}{3\times7}+\frac{2}{7\times11}+\frac{2}{11\times15}+...+\frac{2}{99\times103}\)

\(2\times A=\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{99\times103}\)

\(=\frac{7-3}{3\times7}+\frac{11-7}{7\times11}+\frac{15-11}{11\times15}+...+\frac{103-99}{99\times103}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{99}-\frac{1}{103}\)

\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)

\(\Rightarrow A=\frac{50}{309}\)

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

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

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

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

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

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

Mình sửa lại đề một chút nhé.

\(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)

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

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

A = \(\dfrac{4}{3\times7}\) + \(\dfrac{4}{7\times11}\) + \(\dfrac{4}{11\times15}\) + \(\dfrac{4}{15\times19}\) + \(\dfrac{4}{19\times23}\) + \(\dfrac{4}{23\times27}\)

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

SỬa đề: \(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\)

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

=1/3-1/27

=8/27