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

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

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

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

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

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

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

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

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

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

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

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

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

Theo đề bài ta có:

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

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

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

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

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

tôi cũng hỏi bạn

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

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

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

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

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

a) A = 9/8.11 + 9/11.14 + 9/14.17 + ... + 9/73.75

A = 3.(1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + ... + 1/73 - 1/75)

A = 3.(1/8 - 1/75)

A = 3.67/600

A = 67/200

Các bài sau làm tương tự, riêng câu D thì phân tích ra 

Mình chỉ làm hộ bạn câu a) thôi nhé vì đề sàn sàn giống nhau :

a) \(A=\frac{9}{8×11}+\frac{9}{11×14}+\frac{9}{14×17}+...+\frac{9}{73×75}\)

\(A=\frac{9}{8}-\frac{9}{11}+\frac{9}{11}-\frac{9}{14}+\frac{9}{14}-\frac{9}{17}+...+\frac{9}{73}-\frac{9}{75}\)

\(A=\frac{9}{8}-\frac{9}{75}\)

\(A=\frac{675}{600}-\frac{72}{600}\)

\(A=\frac{673}{600}\)

Vậy,...

Cbht

\(d,D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(10D=\frac{1}{7\cdot8}+\frac{1}{2\cdot70}+\frac{1}{2\cdot130}+...+\frac{1}{2\cdot700}\)

\(10D=\frac{1}{7}-\frac{1}{8}+\frac{1}{2}-\frac{1}{70}+\frac{1}{2}-\frac{1}{130}+...+\frac{1}{2}-\frac{1}{700}\)

\(10D=\frac{1}{7}-\frac{1}{700}\)

\(10D=\frac{99}{700}\)

\(D=\frac{99}{700}:10\)

\(D=\frac{99}{7000}\)

gợi ý nè

1/a.b=1/a-1/b

b. B= 1/25×27+1/27×29+1/29×31+...+1/73×75

2B= 2.(1/25×27+1/27×29+1/29×31+...+1/73×75)

2B= 2/25×27+2/27×29+2/29×31+...+2/73×75

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

2B= 1/25-1/75

B=(1/25-1/75):2

Từng đó đã nha để mik nghĩ tiếp

T.i.c.k đc ko PLEASE!!!