Vì tử bằng mẫu

Đặt A= 200- (3+\(\frac{2}{3}+\frac{2}{4}+.....+\frac{2}{100}\))

         =\(197-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)

        =\(\frac{197.2}{2}-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)

        =\(2.\left(\frac{196+1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)

        =\(2\left(\frac{196}{2}+\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{100}\right)\)

         =\(2\left(98+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)

         =\(2\left(\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+.....+1-\frac{1}{100}\right)\)

          =\(2\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.....+\frac{99}{100}\right)\)

Khi đó \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=\(\frac{2\left(\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=2(đpcm)

Câu này trong đề kì thi Hà NỘi phải ko

ta có 200-(3+\(\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\)

=\(1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)

=\(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

thay \(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

ta có \(\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\left(dpcm\right)\)

bvcdcvxcx

co cach tra loi khac day

a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)

\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)

\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)

b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)

a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
    = \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
    = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)                                                          
    = \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
    = \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ... 

a) \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(100=\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(100=1+1+1+...+1\)

\(\Rightarrow100=100\)

b) \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

A=(2/3+3/4+...+99/100)x(1/2+2/3+3/4+...+98/99)-(1/2+2/3+...+99/100)x(2/3+3/4+4/5+...98/99)

ta cho nó dài hơn như sau

A=(2/3+3/4+4/5+5/6+....+98/99+99/100)

ta thấy các mẫu số và tử số giống nhau nên chệt tiêu các số

2:3:4:5...99 vậy ta còn các số 2/100

ta làm vậy với(1/2+2/3+3/4+.....+98/99) thi con 1/99

làm vậy với câu (1/2+2/3+...+99/100) thì ra la 1/100

vậy với (2/3+3/4+...+98/99) ra 2/99

xùy ra ta có 2/100.1/99-1/100.2/99=1/50x1/99-1/100x2/99=tự tinh nhe mình ngủ đây