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

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

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

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

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

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

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

cunasai

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

\(\Rightarrow100=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}+1+\frac{1}{2}+...+\frac{1}{100}\)

\(\Rightarrow100=1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)+...+\left(\frac{99}{100}+\frac{1}{100}\right)\)

\(\Rightarrow100=1+1+1+...+1\) (100 chữ số 1)

\(\Rightarrow100=100\)

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

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

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

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

1/2+2/3+...+99/100=1-1/2+1-1/3+...+1-1/100=99-(1/2+1/3+,...+1/100)=100-(1+1/2+...+1/100)

Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)

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\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrowđpcm\)

free ire

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

\(=0+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)

\(\RightarrowĐPCM\)

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

Thì ta cần chứng minh 100 = \(\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\right)\)

Biến đổi 

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

            = 1 + 1 + 1 + ...1 (100 số 1) = 100 = Vế trái (đpcm)

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)