Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
Ta có: \(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+2+...+2\right)+\left(3+3+3+...+3\right)+...+\left(2019+2019\right)+2020\)
Trong đó có: 2020 số 1, 2019 số 2, 2018 số 3,..., 2 số 2019, 1 số 2020
Vậy: \(\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+...+2020\)
\(=1\times2020+2\times2019+3\times2018+...+2020\times1\)
\(\Rightarrow A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+3\times2018+...+2020\times1}\)
\(A=\dfrac{1\times2020+2\times2019+3\times2018+...+2020\times1}{1\times2020+2\times2019+3\times2018+...+2020\times1}=1\)
ta có 1/2*2/3*...*2019/2020
=1*2*3*...*2019/2*3*4*..*2020
=1/2020 (rút gọn các số giống nhau)
Ok em, để olm.vn giúp em nhá:
A = \(\dfrac{1}{2}\):3 + \(\dfrac{1}{3}\):4 + \(\dfrac{1}{4}\):5+...+\(\dfrac{1}{2018}\):2019 + \(\dfrac{1}{2019}\): 2020
A=\(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+..+\dfrac{1}{2018}\times\dfrac{1}{2019}+\dfrac{1}{2019}\times\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+....+ \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1009}{2020}\)
Ta có: 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ... + ( 1 + 2 + 3 +...+ 2020)
= ( 1 + 1 + 1 +... + 1 ) + (2 + 2 +...+ 2 ) + ( 3 + 3+...+ 3 ) + ...+ 2020
Có 2020 số 1 ; 2019 số 2 ; 2018 số 3 ;... ; 1 số 2020
= 2020 x 1 + 2019 x 2 + 2018 x 3 + ... + 2020x 1
=> \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+...+2020\times1}\)
= \(\frac{1\times2020+2\times2019+...+2020\times1}{1\times2020+2\times2019+...+2020\times1}=1\)