50 nha                          

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{97}+\dfrac{1}{3}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{\left(\dfrac{1}{1\cdot99}+\dfrac{1}{99\cdot1}\right)+\left(\dfrac{1}{97\cdot3}+\dfrac{1}{97\cdot3}\right)+...+\left(\dfrac{1}{51\cdot49}+\dfrac{1}{49\cdot51}\right)}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{97\cdot3}+...+\dfrac{100}{49\cdot51}}{\dfrac{2}{1\cdot99}+\dfrac{2}{97\cdot3}+...+\dfrac{2}{51\cdot49}}\)

\(=\dfrac{100\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}{2\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}\)

\(=\dfrac{100}{2}\)

\(=50\)

\(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2}{1.99}+\dfrac{2}{3.97}+...+\dfrac{2}{51.49}}\)

\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...+\dfrac{100}{51.49}}\)

\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{1+99}{1.99}+\dfrac{3+97}{3.97}+...+\dfrac{51+49}{51.49}}\)

\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{\dfrac{1}{99}+1+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{51}+\dfrac{1}{49}}\)

\(Q=\dfrac{50(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99})}{1+\dfrac{1}{3}+...+\dfrac{1}{99}}\)

\(\Rightarrow Q=50\)

ụ

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

Tử số = 1 + 1/3 + 1/5 + ... + 1/97 + 1/99

= (1 + 1/99) + (1/3 + 1/97) + ... + (1/49 + 1/51)

= 100/1.99 + 100/3.97 + ... + 100/49.51

= 100.(1/1.99 + 1/3.97 + ... + 1/49.51)

Mẫu số = 1/1.99 + 1/3.97 + 1/5.95 + ... + 1/97.3 + 1/99.1

= 2.(1/1.99 + 1/3.97 + 1/5.95 + ... + 1/49.51)

=> phân số đề bài cho = 100/2 = 50

                    Ta có :

               \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)

              \(=\frac{\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}{2.\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}\)

             \(=\frac{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}{2.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)

             \(=\frac{100.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)

             \(=\frac{100}{2}=50\)

            Ủng hộ mk nha !!! ^_^

Tính B=(1/2+1/3+1/4+....+1/100)/(99/1+98/2+97/3+...+1/99)

Ta rút gọn phần mẫu:

\(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}\\ =\left(\dfrac{1}{1}+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{99}\right)+\left(\dfrac{1}{5}+\dfrac{1}{99}\right)+...+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+\left(\dfrac{1}{1}+\dfrac{1}{99}\right)\\ =\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}\\ =\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{50}\)

Vậy:

\(Q=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\\ =\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{50}}\\ =50\)

@Nguyễn Huy Tú, @Hoàng Thị Ngọc Anh, @Tuấn Anh Phan Nguyễn, @Hoang Hung Quan, @ngonhuminh, và các bn khác giúp mk với!!