đề :

= 1/100 - (1 / 100.99 +1/99.98 + ...+ 1/3.2 +1/2.1 ) 

=1/100 - (1 /1.2 +1/ 2.3 +...+ 1/ 98.99 +1 / 99.100)

=1/100 -( 1- 1/ 2 +1/2 -1/3 +...+1/98 -1/99 +1/99 -1/100)

=1/100 - ( 1- 1/100)

=1/100 - 99 /100

= -98/100

= -49 /50

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

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

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

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

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

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

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

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

=1/100-(1/1x2+1/2x3+...+1/99x100)

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

=1/100-(1-1/100)

=1/100-1+1/100

=2/100-1

=-49/50

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

dễ mà bạn

Vậy bạn làm như thế nào? Chỉ mình đi

k đi đã

chắc chắn làm

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

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

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

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

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

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

\(M=\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)

cộng vào mỗi phân số trong 98 phân số sau,trừ phân số cuối đi 98 , ta được :

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

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

chuyển phân số \(\frac{100}{100}\)ra sau , ta được :

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

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

\(\Rightarrow\frac{M}{N}=\frac{100.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}}=100\)

Thank bn na !!!