A = 2/3*5 + 2/5*7 + 2/7*9 + ... + 2/97*99

A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99

A = 1/3 - 1/99

A = 32/99

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(A=\frac{1}{3}-\frac{1}{99}\)

\(A=\frac{32}{99}\)

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(A=\frac{1}{3}-\frac{1}{99}\)

\(A=\frac{32}{99}\)

đề có sai ko nhỉ ???

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

= \(\frac{1}{3}-\frac{1}{99}\)

= \(\frac{33}{99}-\frac{1}{99}\)

= \(\frac{32}{99}\)

CHO MIK NHA

Nguyễn Lê Bảo An

Hình như sai rồi!

Trên T/S là 2 xog biến tấu 1 hồi mất số 2 của đề cho rồi.

Phải bỏ dòng thứ 2 vào trog ngoặc rồi ở ngoài x vs 2

Kết quả là: 66/99

# Hình như thui nha. Nếu Milk sai thì thôi nha

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

học tốt!!!

=3/2x( 1/3-1/5+1/5-1/7-1/7+....+ 1/97-1/99)

=3/2x( 1/3-1/99)

=3/2x 32/99

= 16/33

a) \(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

       \(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)

       \(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)

       \(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)

       \(=7.\frac{13}{28}=\frac{7.13}{28}=\frac{13}{4}\)

b) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)

      \(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

      \(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

       \(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)

       \(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)

mình cũng làm như trên

\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(E=\frac{1}{1}-\frac{1}{99}\)

\(E=\frac{98}{99}\)

E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt 

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

8D= 1.3.5.8 - 3.5.7.8 + 5.7.9.8 - ... + 97.99.101.8

8D=1.3.5.(7+1)-3.5.7.(9-1)+5.7.9.(11-3) - ... + 97.99.101.(103-95)

8D=1.3.5.7+3.5-3.5.7.9-1.3.5.7+5.7.9.11-3.5.7.9-...+97.99.101.103-95.97.99.101

8D=3.5+97.99.101.103=99900024

D=12487503