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\(\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019.2020-2.2019}{2019\left(2017+1\right)}=\frac{2019\left(2020-2\right)}{2019.2018}=\frac{2019.2018}{2019.2018}=1\)
\(A=\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019\left(2020-2\right)}{2019\left(2017+1\right)}\)
\(=\frac{2019.2018}{2019.2018}=1\)
Vậy \(A=1.\)
Mà lớp 5 làm gì đã học đến dấu \(.\)(dấu nhân lớp 5 viết kiểu này cơ: x )
Chúc em học tốt.
2020 x 2021 - 3031 = 2020 x ( 2 + 2019 ) - 3031 = 2020 x 2019 + 2020 x 2 - 3031 = 2019 x 2020 + 1009
Nên ( 2019 x 2020 + 1009 ) : ( 2020 x 2021 - 3031 ) = ( 2019 x 2020 + 1009 ):( 2019 x 2020 + 1009 )=1
\(A=\frac{2}{2018\times2020}+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{2020-2018}{2018\times2020}+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{1}{2018}-\frac{1}{2020}+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\left(\frac{2021}{2020}-\frac{1}{2020}\right)-\left(\frac{2019}{2018}-\frac{1}{2018}\right)\)
\(A=1-1=0\)
=(2015/ 2019 + 3/2019 + 1/2019 ) : 1/2
= 2019/2019 x 2
= 1 x2
=2
2015/2019:1/2+3/2019:1/2+1/2019:1/2
=(2015/2019+3/2019+1/2019):1/2
=1:1/2
=2
k cho mink nha
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
=> \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+...+2020\times1}\)
= \(\frac{1\times2020+2\times2019+...+2020\times1}{1\times2020+2\times2019+...+2020\times1}=1\)
\(=\dfrac{2019\cdot2020\left(2020\cdot10001-2019\cdot10001\right)}{2020\cdot2019\cdot2018\cdot10001}=\dfrac{10001\cdot1}{10001\cdot2018}=\dfrac{1}{2018}\)