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c) \(M=\frac{2019}{2020}+\frac{2020}{2021}\) và \(N=\frac{2019+2020}{2020+2021}\)
Ta có \(\frac{2019}{2020}>\frac{2019}{2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2020+2021}\)
\(\Rightarrow\frac{2019}{2020}+\frac{2020}{2021}< \frac{2019+2020}{2020+2021}=N\)
\(\Rightarrow M>N\)
\(\dfrac{8}{13}-\dfrac{7}{29}-\dfrac{8}{13}+\dfrac{7}{29}+\dfrac{2019}{2020}=\left(\dfrac{8}{13}-\dfrac{8}{13}\right)-\left(\dfrac{7}{29}+\dfrac{7}{29}\right)+\dfrac{2019}{2020}=0-0+\dfrac{2019}{2020}=\dfrac{2019}{2020}\)
S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2018 – 2019 - 2020 + 2021 + 2022
S = (1 + 2 - 3 - 4) + ... + (2017 + 2018 – 2019 - 2020) + 2021 + 2022
S = (-4) + ... + (-4) + 2021 + 2022
2020 : 4 = 505
S = (-4) . 505 + 2021 + 2022
S = (-2020) + 2021 + 2022
S = 2023
\(1-2+3-4+5-6+...+2019-2020+2021\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2019-2020\right)+2021\)
\(=-1-1-1-..-1+2021\)
\(=-1\cdot1010+2021\)
\(=-1010+2021\)
\(=-1011\)
Ta có : 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... + 2019 - 2020
= ( 1 - 2) + ( 3 - 4 )+ ( 5 - 6 ) + ( 7 - 8 ) + ... + ( 2019 - 2020)
= ( - 1 ) + ( -1 ) + ( - 1 ) + ( - 1 ) + .... + ( - 1 )
\---------------------------------------------------------------/
có tất cả 1010 số ( - 1 )
= ( - 1 ) . 1010
= ( - 1010 )
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
số năm nay luôn
x + ( -2019) = ( -2019 ) + 2020
x = 2020 [cộng hai bên cho ( -2019 ) ]
Vậy x = 2020
\(\left(8^{2020}+8^{2019}\right):8^{2019}\)
\(=\frac{8^{2020}+8^{2019}}{8^{2019}}\)
\(=\frac{8^{2019}\cdot\left(8^1+1\right)}{8^{2019}}\)
\(=8^1+1\)
\(=8+1\)
\(=9\)