1/2.3+1/3.4+1/4.5+...+1/99.100

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

=\(\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

=1/2-1/3+1/3-1/4+1/4-1/5+......+1/99-1/100

=1/2-1/100

=49/100

\(\left(1+\frac{1}{2.3}\right)\left(1+\frac{1}{3.4}\right)\left(1+\frac{1}{4.5}\right)...\left(1+\frac{1}{99.100}\right)\)

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

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

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

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

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

tới đây bí 

\(S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(S=\frac{1}{2}-\frac{1}{100}\)

\(S=\frac{49}{100}\)

chúc các bạn học tốt

\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(S=1\times\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(S=1\times\frac{49}{100}\)

\(S=\frac{49}{100}\)

Ta có: \(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(S=\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{50-1}{100}=\frac{49}{100}\)

Vậy \(S=\frac{49}{100}\)

hi bn

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

Ai tích cho toi thi h lai cho

Ta có: 1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
         = 1/2-1/100
         = 50/100-1/100
         = 49/100

A = 1/2 - 1/3 + 1/3 -1/4 + 1/4 -1/5 + ...+ 1/98 - 1/99 + 1/99 - 1/100 
Ta thấy đoạn giữa sẽ bị trừ lẫn nhau nên bằng 0 
A = 1/2 - 1/100 = 49/100

tích nha

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{100}\)

\(=\dfrac{49}{100}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)

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

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

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{50}{100}-\frac{1}{100}\)

\(=\frac{49}{100}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{99.100}\)

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

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

S=1-1/100=99/100

bạn tách ra, 1/1.2=1-1/2 cứ như thế, rồi trừ đi còn 1-1/100=99/100

S = 99/100

Bài 1 :

Đặt A=1.2+2.3+3.4+4.5+.........+99.100

=> 3A=1.2.3+2.3.(4-1)+........+99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+........+99.100.101-98.99.100

3A=99.100.101

A=33.100.101

A=333300

Bài 2 :

1:20 + 1:44 + 1:77 + 1:119 + 1:170 = \(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}=\frac{1}{10}=0,1\)

           1)1.2+2.3+3.4+4.5+...+99.100

          đặt 3D=1.2+2.3+3.4+...+99.100

                   =1.2.3+2.8.3+...+3.4.3+4.5.3+...+99.100.3

                   =1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

                  =1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5

                  =99.100.101

                 =999900

              D=999900:3=333300 

nếu đúng nhớ cảm ơn nhak. mình ko bít làm bài 2

\(\frac{1}{20}+\left(\frac{1}{44}+\frac{1}{77}\right)+\left(\frac{1}{119}+\frac{1}{170}\right)=\frac{1}{20}+\left(\frac{1}{11}.\frac{1}{4}+\frac{1}{11}.\frac{1}{7}\right)+\left(\frac{1}{17}.\frac{1}{7}+\frac{1}{17}.\frac{1}{10}\right)\)

= \(\frac{1}{20}+\frac{1}{11}.\left(\frac{1}{4}+\frac{1}{7}\right)+\frac{1}{17}.\left(\frac{1}{7}+\frac{1}{10}\right)=\frac{1}{20}+\frac{1}{11}.\frac{11}{28}+\frac{1}{17}.\frac{17}{70}=\frac{1}{20}+\frac{1}{28}+\frac{1}{70}\)

= \(\frac{1}{20}+\frac{1}{14}.\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{1}{20}+\frac{1}{14}.\frac{7}{10}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=0,1\)