(1 + 1/1.3) (1 + 1/2.4) (1+ 1/3.5) ... (1 + 1/2019.2021)
giúp me
(1+1/1.3) . (1+1/2.4) . (1+1/3.5) .... (1+1/2019.2021
Tính :
(1 + \(\dfrac{1}{1.3}\) ) . ( 1+\(\dfrac{1}{2.4}\) ) . (1+\(\dfrac{1}{3.5}\)) . ... . ( 1+\(\dfrac{1}{2019.2021}\))
Lời giải:
Gọi tích trên là $A$
Xét thừa số tổng quát: $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Thay $n=1,2,3....,2019$ ta có:
$A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2020^2}{2019.2021}$
$=\frac{2^2.3^2...2020^2}{(1.3)(2.4)(3.5)...(2019.2021)}$
$=\frac{(2.3....2020)(2.3...2020)}{(1.2.3...2019)(3.4...2021)}$
$=2020.\frac{2}{2021}=\frac{4040}{2021}$
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
Tính giá trị của biểu thức :\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2019.2021}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)
\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)
\(=\dfrac{1}{2021}\)
\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)
Thực hiện phép tính (1+1/1.3)(1+1/2/4)(1+1/3.5).....(1+1/2019.2021)
Giúp mình với ngày mai mình phải nộp rồi!
1/1.3 + 1/3.5 + 1/5.7 +...+ 1/2019.2021
Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2019\cdot2021}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2019\cdot2021}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2019}-\frac{1}{2021}\)
\(2A=1-\frac{1}{2021}=\frac{2020}{2021}\)
\(A=\frac{2020}{2021}:2=\frac{2020\cdot2}{2021}=\frac{4040}{2021}\)
bn
Tran Le Khanh Linh lm sai r nếu chia 2 thì 2021.2 chứ ko phải 2020.2
c, C = 2020/1.2 + 2020/2.3 + 2020/3.4 + ... + 2020/2019.2020
d, D = 2020/1.3 + 2020/3.5 + 2020/5.7 + ... + 2020/2019.2021
e, E = 2023/ 1.3 + 2023/3.5 + 2023/5.7 + ... + 2023/2019.2020
f, F = 1/15 + 1/35 + 1/63 + ... + 1/657
giúp với mình cần gấp lắm
Tính giá trị của biểu thức:
P=\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2020.2022}\right)\)
HELP ME!
P = (1 + \(\dfrac{1}{1.3}\)).(1 + \(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))....(1 + \(\dfrac{1}{2020.2022}\))
P = \(\dfrac{1.3+1}{1.3}\). \(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)....\(\dfrac{2020.2022+1}{2020.2022}\)
P=\(\dfrac{\left(2-1\right)\left(2+1\right)+1}{1.3}\).\(\dfrac{\left(3-1\right)\left(3+1\right)+1}{2.4}\)...\(\dfrac{\left(2021+1\right).\left(2022-1\right)+1}{2020.2022}\)
P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)....\(\dfrac{2021.2021}{2020.2022}\)
P = \(\dfrac{2.2021}{2022}\)
P = \(\dfrac{2021}{1011}\)
Tính gía trị của biểu thức:
\(P=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2020.2022}\right)\)
HELP ME!
P = (1+\(\dfrac{1}{1.3}\)).(1+\(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))...(1+\(\dfrac{1}{2020.2022}\))
P =\(\dfrac{1.3+1}{1.3}\).\(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)...\(\dfrac{2020.2022+1}{2020.2022}\)
P = \(\dfrac{(2-1)(2+1)+1}{1.3}\).\(\dfrac{(3-1)(3+1)+1}{2.4}\)...\(\dfrac{(2021-1)(2021+1)}{2020.2022}\)
P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)...\(\dfrac{2021.2021}{2020.2022}\)
P = \(\dfrac{2021}{1011}\)