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

3A= 3.(1.2 + 2.3 + 3.4 + ..... +99.100)

3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)

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

3A=99.100.101

A=99.100.101/3=333300

đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

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

=>3A=1.2.3+2.3.3+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+...+99.100.101‐98.99.100

=1.2.3‐1.2.3+2.3.4‐2.3.4+3.4.5‐3.4.5+4.5.6‐4.5.6+...+99.100.101

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300

đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

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

=>3A=1.2.3+2.3.3+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+...+99.100.101‐98.99.100

=1.2.3‐1.2.3+2.3.4‐2.3.4+3.4.5‐3.4.5+4.5.6‐4.5.6+...+99.100.101

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300

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

3A=1.2.(3-0) + 2.3.(4-1) +.....+99.100.(101-98)

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

3A=99.100.101

A=99.100.101/3=333300

=3080 nhớ tít

=> 3A = 3 [ 1.2 + 2.3 + 3.4 + ... + (n-1).n ]

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 +... + 1001.1002.3 

=> 3A =  1.2.3 + 2.3 . ( 4-1 ) +3.4.( 5-2 ) + ... + 1001.1002 ( 1003-1000 )

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +... + 1001.1002 .1003 - 1000.1001.1002

=> 3A = 1001.1002.1003 

=> A = 1001 . 1002 . 1003 : 3 

=> A = ?

cái nay có trong sách bạn ak

chữ số tận cùng là 8

\(\text{#}HaimeeOkk\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)

\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)

\(A=1-\dfrac{1}{2020}\)

\(A=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

\(1+\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+\frac{7}{3\cdot4}+...+\frac{7}{59\cdot60}\)

\(=1+7\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{59\cdot60}\right)\)

\(=1+7\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=1+7\left(1-\frac{1}{60}\right)\)

\(=1+7\cdot\frac{59}{60}\)

Cảm ơn bạn nha Sooya.

\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)

     \(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

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

      \(=5.\left(1-\frac{1}{100}\right)\)

   \(=5.\frac{99}{100}\)

      \(=\frac{99}{20}\)

5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100

= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )

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

= 5 . ( 1 - 1/100 )

= 5 . 99/100

= 99/20

\(\frac{99}{100}\)nhé

Xét: 3/(1.2)=(1+2)/(1.2)=1/2+1

5/(2.3)=(2+3)/(2.3)=1/3+/1/2

7/(3.4)=(3+4)/(3.4)=1/4+1/3

...

2021/(1010.1011)=(1010+1011)/(1010.1011)=1/1010+1/1011

Do đó: Q=1+1/2-1/2-1/3+1/3+1/4-...-1/1010-1/1011

=1-1/1011

=1010/1011

sai thì sorry nha, nhưng cách làm là đúng rồi

Lê Gia Long e lớp 4 thì ko đc lm!

\(Q=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}...-\frac{2021}{1010\cdot1011}\)

\(=\left(\frac{1}{1}+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)...-\left(\frac{1}{1010}-\frac{1}{1011}\right)\)

\(=1-\frac{1}{1011}\)

\(=\frac{1010}{1011}\)

mik mơí lớp 4

 hihi

I don't now

mik ko biết 

sorry 

......................

b,\(B=2^2+4^2+...+20^2\)

\(\Rightarrow B=2^2\left(1^2+2^2+...+10^2\right)\)

\(\Rightarrow B=4.\left[1.\left(2-1\right)+2.\left(3-1\right)+...+10.\left(11-1\right)\right]\)

\(\Rightarrow B=4\left(1.2-1+2.3-2+...+10.11-10\right)\)

\(\Rightarrow B=4\left[\left(1.2+2.3+...+10.11\right)-\left(1+2+...+10\right)\right]\)

\(\Rightarrow B=4\left(\frac{10.11.12}{3}-\frac{11.10}{2}\right)\)

c)C=(1.2+2.3+3.4+...+20.21)-(1²+2²+3²+...+20²)

=>C=1.2+2.3+3.4+...+20.21-1.1-2.2-3.3-...-20.20

=>C=1.(2-1)+2.(3-2)+3.(4-3)+...+20.(21-20)

=>C=1+2+3+...+20

=>C=(1+20)+(2+19)+...+(15+16)

=>C=21+21+...+21

=>C=21.10

=>C=210