tìm x nguyên 2023+2022+2021+2020+...+x=2023
2022 x 2023 - 3 / 2023 x 2021 + 2020
cứu mình với
Lời giải:
\(\frac{2022\times 2023-3}{2023\times 2021+2020}=\frac{2023\times (2021+1)-3}{2023\times 2021+2020}
\\
=\frac{2023\times 2021+2023-3}{2023\times 2021+2020}=\frac{2023\times 2021+2020}{2023\times 2021+2020}=1\)
tìm x
(x+1)/2023 + (x+2)/2022=(x+3)/2021 + (x+4)/2020
\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)
`(x+1)/2023+(x+2)/2022=(x+3)/2021+(x+4)/2020`
`=>(x+1)/2023+1+(x+2)/2022+1=(x+3)/2021+1+(x+4)/2020+1`
`=>(x+2024)/2023+(x+2024)/2022=(x+2024)/2021+(x+2024)/2020`
`=>(x+2024)/2023+(x+2024)/2022-(x+2024)/2021-(x+2024)/2020=0`
`=>(x+2024).(1/2023+1/2022-1/2021-1/2020)=0`
Vì `1/2023+1/2022-1/2021-1/2020` `\ne` `0`
`=> x+2024=0`
`=>x=-2024`
Bài 2: 2023+2022+2021+2020+.....+x=2023 (Trong đó về trái là tổng các số nguyên liên tiếp viết theo thứ tự giảm dần).
Giup mk à:(
2023+2022+2021+....+x=2023
2022+2021+....+x=2023-2023
2022+2021+...+x=0
=> x=-2022-2021-....(khoảng này ko biết kết thúc ở đâu nên ko tính đc)
x - 2021/2020 + x-2021/2021 - x- 2021/2022 - x- 2021/2023= 0
x= 2002/3000
ko bt đúng ko mong bn nhắc nhở
2020 + 2021/2021 + 2022 và 2020 + 2021 + 2022/2023
So sánh:
B=2020/2021+2021/2022+2022/2023+2023/2020 và 4
Giải chi tiết giúp minh với ạ
Mai mình phải nộp rồi
2020/2021<1
2021/2022<1
2022/2023<1
2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023
=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4
Tìm x, biết:
( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2023}\) ) . x = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) + \(\dfrac{2020}{3}\)
+ ... + \(\dfrac{1}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
\(P\left(x\right)\)=\(x^{2023}-2024.x^{2022}+2024.x^{2021}-2024.x^{2020}+.....+2024.x-1\)
tính P ( 2023)
Giải nhanh giúp mik ạ !! đang cânf gấp O(∩_∩)O
Với x = 2023
<=> x + 1 = 2024
Khi đó P(2023) = x2023 - (x + 1).x2022 + ... + (x + 1).x - 1
= x2023 - x2023 - x2022 + .. + x2 + x - 1
= x - 1 = 2023 - 1 = 2022
Cho các số thực x,y,z thỏa mãn x^2022+y^2022+z^2022=x^2023+y^2023+z^2023, tính P=x^2021+y^2022+z^2023.