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\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{75}\)

\(B=\frac{1}{75}\)

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+....+\frac{1}{73.75}\)

\(B=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\cdot\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

Theo đề bài ta có:

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(\Leftrightarrow B=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(\Leftrightarrow B=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}\right)\)

\(\Leftrightarrow B=\frac{1}{2}\cdot\frac{2}{75}\)

\(\Leftrightarrow B=\frac{1}{75}\)

\(\left(\frac{1}{25x27}+\frac{1}{27x29}+\frac{1}{29x31}+....+\frac{1}{73x75}\right)x\frac{3}{10}=x\)

\(< =>\frac{1}{2}x\left(\frac{2}{25x27}+\frac{2}{27x29}+\frac{2}{29x31}+....+\frac{2}{73x75}\right)x\frac{3}{10}=x\)

\(< =>\frac{1}{2}x\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+....+\frac{1}{73}-\frac{1}{75}\right)x\frac{3}{10}=x\)

\(< =>\frac{1}{2}x\left(\frac{1}{25}-\frac{1}{75}\right)x\frac{3}{10}=x< =>\frac{1}{2}x\frac{2}{25}x\frac{3}{10}=x< =>x=\frac{3}{250}\)

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

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

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

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

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 

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

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

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

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

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

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

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

D=6/15.18+6/18.21+...+6/89.92

D=6(1/15.18+1/18.21+...+1/89.92)

a) 3D=6(1/15-1/18+1/18-1/21+...+1/89-1/92)

3D=6(1/15-1/92)

3D=6.(77/1380)

3D=77/230

D=77/690

b) F=1/25.27+1/27.29+...+1/73.75

2F=2/25.27+2/27.29+..+2/73.75

2F=1/25-1/27+1/27-1/29+...+1/73-1/75

2F=1/25-1/75

2F=2/75

F=1/75

a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)

\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)

\(\frac{1}{4}xA=\frac{127}{384}\)

\(A=\frac{127}{384}:\frac{1}{4}\)

\(A=\frac{127}{96}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\\ A=\frac{1}{75}\)

\(B=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146+150}=\frac{1}{4}\left(\frac{15}{90}-\frac{15}{94}+\frac{15}{94}-\frac{15}{98}+...+\frac{15}{146}-\frac{15}{150}\right)\)

\(B=\frac{1}{4}\left(\frac{15}{90}-\frac{15}{150}\right)=\frac{1}{60}\)