1. Tính bằng cách thuận tiện
\(\dfrac{2020\cdot2020\cdot20192019-2019\cdot2019\cdot20202020}{2020\cdot2019\cdot20182018}\)
A=\(\frac{2019\cdot2020-4038}{2017\cdot2019+2019}\)
\(\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019.2020-2.2019}{2019\left(2017+1\right)}=\frac{2019\left(2020-2\right)}{2019.2018}=\frac{2019.2018}{2019.2018}=1\)
\(A=\frac{2019.2020-4038}{2017.2019+2019}\)
\(=\frac{2019\left(2020-2\right)}{2019\left(2017+1\right)}\)
\(=\frac{2019.2018}{2019.2018}=1\)
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.
\(A=\frac{2019.2020-4038}{2017.2019+2019}\)
\(A=\frac{2019.2020-2.2019}{2017.2019+2019}\)
\(A=\frac{2019.\left(2020-2\right)}{2019.\left(2017+1\right)}\)
\(A=\frac{2018}{2018}\)
\(A=1\)
Tham khảo nhé~
\(\frac{2020}{2019}-\frac{2018}{2017}+\frac{2}{2017\cdot2019}3\cdot x-18\)
Tính bằng cách thuận tiện :
( 2019 x 2020 + 1009 ) : ( 2020 x 2021 - 3031 )
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
tính bằng cách thuận tiện
1-2 + 3-4 + 5-6 + 7-8 + ....... + 2019-2020 + 2021
tính tổng sau:
\(A=1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2019\cdot2020^2\)
\(A=1.2.3.4...2019.\left(2020.2021-2020^2\right)=1.2.3.4...2019.2020\)
tính :P=\(\dfrac{\left(2016^2\cdot2026+31\cdot2017-1\right)\left(2016\cdot2021+4\right)}{2017\cdot2018\cdot2019\cdot2020\cdot2021}\)
Đặt \(2016=a\) biểu thức trên trở thành:
\(P=\dfrac{\left(a^2\left(a+10\right)+31\left(a+1\right)-1\right)\left(a\left(a+5\right)+4\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)}=\dfrac{A}{B}\)
Với \(B=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)\)
Ta có: \(a^2\left(a+10\right)+31\left(a+1\right)-1=a^3+10a^2+31a+30\)
\(=a^3+5a^2+6a+5a^2+25a+30=a\left(a^2+5a+6\right)+5\left(a^2+5a+6\right)\)
\(=\left(a+5\right)\left(a^2+5a+6\right)=\left(a+5\right)\left(a^2+2a+3a+6\right)\)
\(=\left(a+5\right)\left(a+2\right)\left(a+3\right)\)
Và \(a\left(a+5\right)+4=a^2+5a+4=a^2+a+4a+4=\left(a+1\right)\left(a+4\right)\)
\(\Rightarrow A=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)=B\)
\(\Rightarrow P=\dfrac{A}{B}=1\)
Tính bằng cách thuận tiện:
A= 2/2018*2020 + 2021/2020 - 2019/2018
\(A=\frac{2}{2018\times2020}+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{2020-2018}{2018\times2020}+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\frac{1}{2018}-\frac{1}{2020}+\frac{2021}{2020}-\frac{2019}{2018}\)
\(A=\left(\frac{2021}{2020}-\frac{1}{2020}\right)-\left(\frac{2019}{2018}-\frac{1}{2018}\right)\)
\(A=1-1=0\)
Tính bằng cách thuận tiện nhất:
2020 : 0,2 + 2020 + 2020 : 0,25
\(2020:0,2+2020+2020:0,25\)
\(=2020:\dfrac{1}{5}+2020+2020:\dfrac{1}{4}\)
\(=2020\times5+2020+2020\times4\)
\(=2020\times \left(5+1+4\right)\)
\(=2020\times10\)
\(=20200\)
Tính giá trị biểu thức \(P=\frac{\left(2016^2\cdot2026+31\cdot2017-1\right)\left(2016\cdot2021+4\right)}{2017\cdot2018\cdot2019\cdot2020\cdot2021}\)