Tính nhanh :
(-2018)-2017+(2018-2020+2021)= ?
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A = 2021/2022+2020/2021+2019/2020+2018/2019+2017/2018
A<2022/2022+2021/2021+2020/2020+2019/2019+2018/2018
A<1+1+1+1+1
A<5
2011+2012+2013+2014+2015+2016+2017+2018+2019+2020+2021+ 2022+2023 =(2011+2023)+(2013+2022)+...+(2016+2018)+2017 =4034+4034+4034+4034+4034+4034+2017 =4034x6+2017=26221
2011+2012+2013+2014+2015+2016+2017+2018+2019+2020+2021+2022+2023
=(2011+2023)+(2013+2022)+...+(2016+2018)+2017 =4034+4034+4034+4034+4034+4034+2017 =4034x6+2017=26221
a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)
= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)
= 1
@Nguyen Thi Ngoc Linh
\(A=1-3+5-7+......-2019+2021-2023\)
\(A=\left(1-3\right)+\left(5-7\right)+....+\left(2021-2023\right)\)
\(A=-2+\left(-2\right)+....+\left(-2\right)\left(506 cặp\right)\)
\(A=-2.506\)
\(A=-1012\)
*) A=(1-3)+(5-7)+....+(2021-2023)
<=> A=-2+(-2)+...+(-2)
Dãy A có (2023-1):2+1=1012 số số hạng
=> Có 506 số (-2)
=> A=(-2).506=-1012
Đáp án: 1
TA CÓ:
E=1+(2-3-4+5)+(6-7-8+9)+.......+(2018-2019-2020+2021)
E=1+0+0+0+.....+0
E=1
K CHO MIK NHAAAAA
\(\frac{2018\cdot2016+2021}{2017\cdot2018+3}\)
\(=\frac{2018\cdot2016+2021}{2016\cdot2018+2018\cdot1+3}\)
\(=\frac{2018\cdot2016+2021}{2016\cdot2018+2021}\)
\(=1\)
\(-2018-2017+\left(2018-2020+2021\right)\)
\(=-2018-2017+2018-2020+2021\)
\(=\left(-2018+2018\right)-\left(2017+2020\right)+2021\)
\(=0-4037+2021\)
\(=-4037+2021\)
\(=-2016\)
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