\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)

\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)

\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(\Rightarrow C=\frac{1}{100}-1+\frac{1}{100}\)

\(\Rightarrow C=\left(\frac{1}{100}+\frac{1}{100}\right)-1\)

\(\Rightarrow C=\frac{1}{50}-1\)

\(\Rightarrow C=\frac{-49}{50}\)

dễ mà bạn

Vậy bạn làm như thế nào? Chỉ mình đi

1-2+3-4+....+2015-2016+2017

=2017-2016+2015-2014+.....+4-3+2-1

=1+1+1+...+1(Có 1013 số 1)

= 1013

Ta có: \(\frac{99}{1}+\frac{98}{2}+\cdots+\frac{1}{99}\)

\(=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+\cdots+\left(1+\frac{1}{99}\right)+1\)

\(=\frac{100}{2}+\frac{100}{3}+\cdots+\frac{100}{99}+\frac{100}{100}\)

\(=100\left(\frac12+\frac13+\cdots+\frac{1}{100}\right)\)

Ta có: \(C=\frac{\frac12+\frac13+\frac14+\cdots+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\cdots+\frac{1}{99}}\)

\(=\frac{\frac12+\frac13+\cdots+\frac{1}{100}}{100\left(\frac12+\frac13+\cdots+\frac{1}{100}\right)}=\frac{1}{100}\)

Xin hãy giúp mình

a,=3/2*4/3*....100/99

=3*4*5*....*100/2*3*...*99

=100/2=50

b, nhân lên băng:

1*2*3*...*99/2*3*...*100=1/100

\(\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}+\frac{99}{1}}\)

Xét M - 99 + 98 = \(\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)

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

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

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