tính nhanh:2²/1.3+3²/2.4+4²/3.5+...+999²/998.1000
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NP
0
LA
6 tháng 3 2018
\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)
\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)
\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)
20 tháng 5 2020
=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????
NH
2
7 tháng 5 2015
\(A=\frac{2^2.3^2.4^2............99^2}{1.3.2.4.3.5................998.1000}\)
\(A=\frac{1.2.3.4.5................999.1.2.3.4................999}{1.2.3.4.5.6.7..........1000.1.2.3.4..............998}\)
\(A=\frac{999.999}{1000.998}\)
\(Ko\) \(\text{chắc lắm}\)
DD
1
15 tháng 1 2018
A = (2.3.4. .... .999/1.2.3. .... .998) . (2.3.4. .... .999/3.4.5. ..... .1000)
= 999. 2/1000
= 999/500
Tk mk nha
5 tháng 4 2018
a, \(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{999.999}{998.1000}\)
\(=\frac{2.3.4...999}{1.2.3...998}.\frac{2.3.4...999}{3.4.5...1000}\)
\(=\frac{999}{1}.\frac{2}{1000}\)
\(=\frac{999.2}{1000.1}=\frac{999.2}{500.2.1}\)
\(=\frac{999}{500}\)
Vậy \(A=\frac{999}{500}\)
chúc bạn học giỏi
LT
1
MV
15 tháng 1 2018
\(A=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{999^2}{998\cdot1000}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot999^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot998\cdot1000}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot999\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot999\right)}{\left(1\cdot2\cdot3\cdot...\cdot998\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot1000\right)}\\ =\dfrac{2\cdot3\cdot4\cdot...\cdot999}{1\cdot2\cdot3\cdot...\cdot998}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot999}{3\cdot4\cdot5\cdot...\cdot1000}\\ =999\cdot\dfrac{1}{500}\\ =\dfrac{999}{500}\)
17 tháng 1 2018
\(I=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)
\(I=\frac{2^2.3^2.4^2...999^2}{2.3^2.4^2...998^2.1000}\)
\(I=\frac{2}{1000}=\frac{1}{500}\)
17 tháng 1 2018
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{999\cdot999}{998\cdot1000}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot999\cdot999}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot998\cdot1000}\)
\(=\frac{2\cdot3\cdot4\cdot...\cdot999}{1\cdot2\cdot3\cdot...\cdot998}\cdot\frac{2\cdot3\cdot4\cdot...\cdot999}{3\cdot4\cdot5\cdot...\cdot1000}\)
\(=\frac{999}{1}\cdot\frac{2}{1000}\)
\(=\frac{999}{500}\)
VT
0
\(S=\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{998.1000}\right)=\)
\(=998+\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{997.999}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{998.1000}\right)\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{997.999}\)
\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{997.999}=1-\dfrac{1}{999}=\dfrac{998}{999}\)
\(\Rightarrow A=\dfrac{488}{999}\)
\(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{998.1000}\)
\(\Rightarrow2B=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{998.1000}=\dfrac{1}{2}-\dfrac{1}{1000}=\dfrac{499}{1000}\)
\(\Rightarrow B=\dfrac{499}{2000}\)
\(\Rightarrow S=998+\dfrac{499}{999}+\dfrac{499}{2000}\)