a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

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

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

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

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

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

 = \(1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(=\frac{99}{100}\)

thấy đúng thì k cho mk nha mấy bạn

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(=\frac{99}{100}\)

b: Tổng của N là:

\(\dfrac{49\cdot48}{2}=49\cdot24=1176\)

chào nick thứ 2 đây

a) \(3M=1.2.3+2.3.3+...+48.49.3=1.2.3+2.3.\left(4-1\right)+...+48.49.\left(50-47\right)=1.2.3+2.3.4-1.2.3+...+48.49.50-47.48.49=48.49.50\Rightarrow M=\dfrac{48.49.50}{3}\Rightarrow M=39200\)

b) Tương tự câu a

v

Answer:

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

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

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

\(=\dfrac{99}{100}\)

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

=1-1/00

=99/100

Tick

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

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

     \(=1-\frac{1}{100}=\frac{99}{100}\)

\(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}-...-\frac{1}{99\cdot100}\)

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

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

=\(\frac{99}{100}\)