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 \(A=50\)

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

\(=2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)

\(=2-\frac{1}{99}\)

\(=\frac{197}{99}\)

1+1/3+1/5...+1/97+1/99=(1+1/99) + (1/3+1/97) + (1/5+1/95)....+(1/49+1/51) 
= 100/1.99 + 100/3.97 + 100/5.95 +.....=100.(1/1.99 + 1/3.97 + 1/5.95 +.....) 
Mau so: 
1/1.99 + 1/3.97 +1/5.95....+1/95.5+ 1/97.3 +1/99.1=2/1.99 +2/3.97 +2/5.95+..... 
=2.(1/1.99 + 1/3.97 + 1/5.95 +.....) 
=>A=(100.(1/1.99 + 1/3.97 + 1/5.95 +.....)) : (2.(1/1.99 + 1/3.97 + 1/5.95 +.....))=50

chuẩn luôn , tích nha

Thanks nhìu  ^_^

đúng ko hoài mình cho bạn nhé

ai tích mình mình tích lại

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)

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

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

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

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

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

\(=50\)