[3.4]+[-4.5]Giúp mình với
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\(C=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{2013.2016}{2014.2015}\)
\(C=\frac{\left(1.2.3.....2013\right).\left(4.5.6.....2016\right)}{\left(2.3.4.....2014\right).\left(3.4.5.....2015\right)}\)
\(C=\frac{1}{2014}.\frac{2016}{3}\)
\(C=\frac{336}{1007}\)
A=3/1.2+3/2.3+3/3.4+3/4.5+...+3/2021.2022
A=3(1/1.2+1/2.3+1/3.4+1/4.5+...+1/2021.2022)
A=3(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2021-1/2022)
A=3[1/1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/2021-1/2021)-1/2022]
A=3[1/1+0+0+0+...+0-1/2022
A=3(1/1-1/2022)
A=3(2022/2022-1/2022)
A=3.2021/2022
A=2021/674
Bn Tham Khảo:
https://hoc247.net/hoi-dap/toan-6/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-2016-faq188428.html
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
Dạng tổng quát :
\(\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{n\left(n+1\right)}=\frac{1}{2}-\frac{1}{n+1}=\frac{n+1-2}{2\left(n+1\right)}=\frac{n-1}{2\left(n+1\right)}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}\)
\(=\frac{499}{2000}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}+\frac{1}{2017\cdot2018}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)
\(=2-\frac{1}{2018}\)
\(=\frac{1009}{2018}-\frac{1}{2018}\)
\(=\frac{1008}{2018}=\)TỰ RÚT GỌN NHA
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=2-\frac{2007}{2008}\)
\(=\frac{2009}{2008}\)
~Học tốt~
\(\text{#}HaimeeOkk\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)
\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)
\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)
\(A=1-\dfrac{1}{2020}\)
\(A=\dfrac{2019}{2020}\)
Vậy \(A=\dfrac{2019}{2020}\)
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 + 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
tả lời
( 3 . 4 ) + ( -4 . 5 )
= 12 + (-20)
= -8
hok tốt nha bn
ủa ? hỏi j dễ z bẹn ?