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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 5 2023
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
26 tháng 1 2023
Đặt D=2^2021+2^2020+...+2+1
=>2D=2^2022+2^2021+...+2^2+2
=>D=2^2022-1
=>C=2^2022-2^2022+1=1
LP
2
24 tháng 2 2022
\(2A=\dfrac{2^{2021}-1-1}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2B=\dfrac{2^{2022}-1-1}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
mà \(2^{2021}-1< 2^{2022}-1\)
nên A<B
27 tháng 2 2022
A=22020−122021−1A=22020-122021-1
⇒2A=2.(22020−1)22021−1⇒2A=2.(22020-1)22021-1
⇒2A=22021−222021−1⇒2A=22021-222021-1
⇒2A=22021−1−122021−1⇒2A=22021-1-122021-1
⇒2A=1−122021−1⇒2A=1-122021-1
B=22021−122022−1B=22021-122022-1
⇒2B=2.(22021−1)22022−1⇒2B=2.(22021-1)22022-1
21 tháng 8 2019
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\Rightarrow-\frac{1}{12}\left(x-2018\right)=0\Leftrightarrow x=2018\)
B
4 tháng 9 2020
Bài làm :
Ta có :
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\text{Vì : }\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\ne0\Rightarrow x-2018=0\)
\(\Rightarrow x=2018\)
Vậy x=2018
DD
Đoàn Đức Hà
Giáo viên
2 tháng 3 2021
a) \(M=2020+2020^2+...+2020^{10}\)
\(M=\left(2020+2020^2\right)+\left(2020^3+2020^4\right)+...+\left(2020^9+2020^{10}\right)\)
\(M=2020\left(1+2020\right)+2020^3\left(1+2020\right)+...+2020^9\left(1+2020\right)\)
\(M=2021\left(2020+2020^3+...+2020^9\right)⋮2021\).
b) Bạn làm tương tự câu a).
2 tháng 3 2021
b, \(A=2021+2021^2+...+2021^{2020}\)
\(=2021\left(1+2021\right)+...+2021^{2019}\left(1+2021\right)\)
\(=2022\left(2021+...+2021^{2019}\right)⋮2022\)
Vậy ta có đpcm
A=20222−20212+20202−20192+...+22−1�=20222-20212+20202-20192+...+22-1
=(2022−2021)(2022+2021)+(2020−2019)(2020+2019)+....+(2−1)(2+1)=(2022-2021)(2022+2021)+(2020-2019)(2020+2019)+....+(2-1)(2+1)
=1.4043+1.4039+.....+1.3=1.4043+1.4039+.....+1.3
Vì từ 3→40433→4043 có 10111011 số
⇒A=(4043+3).10112⇒�=(4043+3).10112
=4046.10112=2023.1011=4046.10112=2023.1011
=2045253