Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\) (đpcm)

*đpcm = điều phải chứng minh

\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)

\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{8}\right)-\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)

Ta có:

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)

1/1 - 1/ 50 = 49 / 50

**** mình nhé bạn

Ta biến đổi vế phải :

1-1/2+1/3-1/4+.....+1/49-1/50

=(1+1/3+1/5+....+1/49)-(1/2+1/4+1/6+.......+1/50)

=(1+1/2+1/3+.....+1/49+1/50)-2(1/2+1/4+1/6+......+1/50)

=(1+1/2+...+1/50)-(1+1/2+1/3+....+1/25)

=1/26+1/27+.......+1/50

Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50

Mình không bấm phân số được mong mấy bạn thông cảm

\(=\frac{49}{50}\)

\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(< \frac{1}{26}+\frac{1}{26}+\frac{1}{26}+...+\frac{1}{26}+\frac{1}{26}\)
\(=\frac{25}{26}< 1\)(sai với đề bài)

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{60}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)

Đương nhiên là a<b rồi,vì A thuộc B mà

ChoA=1/26+1/27+1/28+..  +1/49, B=1-1/2+1/3-1/4+... +1/49-1/50

B = 1 + 1/2 + 1/3 + ... + 1/50

b = (1 + 1/3 + 1/5 + ... + 1/49) + (1/2 + 1/4 + 1/6 + ... + 1/50)

b = (1 + 1/2 + 1/3 + 1/4 + ... + 1/50) - 2(1/2 + 1/4 + 1/6 + ... + 1/50)

b = 1 + 1/2 + ... + 1/50 - 1 - 1/2 - 1/3 - ... - 1/25

b = 1/26 + 1/27 + 1/28  + ... + 1/50

vậy a = b