( 1/2019 + 2011/2020 + 4012/2021) x (1/2 - 1/3-1/6 )
help meeeeee............
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Nếu 1/3 + 1/6 +1/10 + ...... + 1/x.(x+1) : 2 = 2019/2021
A.x = 2019/2020 B. x = 2019 C. x = 2020 D. x = 2021
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1+2-3-4+5+6-7-8-....-2019-2020+2021+2022 help
Ta có: 1+2-3-4+5+6-7-8+.....-2019-2020+2021+2022
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
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Tính : S = \(1-\dfrac{1}{2}+\dfrac{1}{3}-\)\(\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2021}\)và
P = \(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+...+\dfrac{1}{2020}+\dfrac{1}{2021}\)
Tính : \(\left(S-P\right)^{2022}\)
S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)
= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)
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Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Bài 1: Tính nhanh
A = 1 - 3 + 5 - 7+...- 2019 + 2021 - 2023
B = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 +...+ 2017 + 2018 + - 2019 - 2020
Help me :v
\(A=1-3+5-7+......-2019+2021-2023\)
\(A=\left(1-3\right)+\left(5-7\right)+....+\left(2021-2023\right)\)
\(A=-2+\left(-2\right)+....+\left(-2\right)\left(506 cặp\right)\)
\(A=-2.506\)
\(A=-1012\)
*) 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
\(B=1+2-3-4+5+6-7-8+......+2017+2018-2019-2020\)
\(B=\left(1+2-3-4\right)+\left(5+6-7-8\right)+.....+\left(2017+2018-2019-2020\right)\)
\(B=-4+\left(-4\right)+.....+\left(-4\right)\left(505 cặp\right)\)
\(B=-4.505\)
\(B=-2020\)
Câu 24: Cho biểu thức: A=1/2+1/3+1/4+.........+1/2021+1/2022 Và B=2021/1+2020/2+2019/3+.........+3/2019+2020+1/2021
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
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Toán 6:
Không dùng máy tính hãy so sánh A= 5^2020+1/5^2021+1
và B=10^2019+1/10^2020+1
help mik dc ko ;-;
ta có :
A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)
B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)
\(\Leftrightarrow\) B < A
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Cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}+....+\frac{1}{2019}-\frac{1}{2020}+\frac{1}{2021}\)
Và \(P=\frac{1}{2011}+\frac{1}{2012}+...+\frac{1}{2020}+\frac{1}{2021}\)
Tính \(\left(S-P\right)^{2020}\)
