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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
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
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 + 2.3 + 3.4 + ...... + 99.100
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
\(\text{#}HaimeeOkk\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)
\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)
\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)
\(A=1-\dfrac{1}{2020}\)
\(A=\dfrac{2019}{2020}\)
Vậy \(A=\dfrac{2019}{2020}\)
Sửa đề : `P=3/1.2+3/2.3+3/3.4+....+3/11.12`
`P=3/1.2+3/2.3+3/3.4+....+3/11.12`
`=3(1/1.2+1/2.3+1/3.4+...+1/11.12)`
`=3(1/1-1/2+1/2-1/3+1/3-1/4+...+1/11-1/12)`
`=3(1/1-1/12)`
`=3(12/12-1/12)`
`=3 . 11/12`
`=33/12`
`=11/4`
Vậy `P=11/4`
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
hình đề bị sai thì phải
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+\dfrac{3}{3\cdot4}+...+\dfrac{3}{11\cdot12}\) đề phải ntn chứ nhỉ?
\(=3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{11\cdot12}\right)\)
\(=3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(=3\left(\dfrac{1}{1}-\dfrac{1}{12}\right)\)
\(=3\left(\dfrac{12}{12}-\dfrac{1}{12}\right)\\ =3\cdot\dfrac{11}{12}\\ =\dfrac{33}{12}\\ =\dfrac{11}{4}\)
I don't now
mik ko biết
sorry
......................
b,\(B=2^2+4^2+...+20^2\)
\(\Rightarrow B=2^2\left(1^2+2^2+...+10^2\right)\)
\(\Rightarrow B=4.\left[1.\left(2-1\right)+2.\left(3-1\right)+...+10.\left(11-1\right)\right]\)
\(\Rightarrow B=4\left(1.2-1+2.3-2+...+10.11-10\right)\)
\(\Rightarrow B=4\left[\left(1.2+2.3+...+10.11\right)-\left(1+2+...+10\right)\right]\)
\(\Rightarrow B=4\left(\frac{10.11.12}{3}-\frac{11.10}{2}\right)\)
=> 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 = ?
\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
\(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
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
\(1+\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+\frac{7}{3\cdot4}+...+\frac{7}{59\cdot60}\)
\(=1+7\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{59\cdot60}\right)\)
\(=1+7\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(=1+7\left(1-\frac{1}{60}\right)\)
\(=1+7\cdot\frac{59}{60}\)
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