S=(1-2/2*3)(1-2/3*4)(1-2/4*5)...(1-2/2020*2021)
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
7 tháng 9 2021
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
15 tháng 6 2023
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
ta có :
\(1-\frac{2}{2.3}=\frac{2.3-2}{2.3}=\frac{1.2}{2.3}\)
tương tự : \(1-\frac{2}{3.4}=\frac{2.3}{3.4},....,1-\frac{2}{2020.2021}=\frac{2019.2020}{2020.2021}\)
Vậy \(S=\frac{1.2}{2.3}.\frac{2.3}{3.4}.....\frac{2019.2020}{2020.2021}=\frac{1.\left(2.3...2019\right)^2.2020}{2.\left(3.4....2020\right)^2.2021}=\frac{2}{2020.2021}\)