2. tìm x
\(\dfrac{x-4}{2020}\)+\(\dfrac{x-3}{2021}\)+\(\dfrac{x-2}{2022}\)+\(\dfrac{x-1}{2023}\)+\(\dfrac{x-2024}{5}\)=2
K
Khách
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Những câu hỏi liên quan
HK
16 tháng 4 2023
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). 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
BP
3
NH
3 tháng 4 2023
\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
`<=>(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)(1/2023+1/2022-1/2021-1/2020)=0`
`<=>x-2024=0(1/2023+1/2022-1/2021-1/2020>0)`
`<=>x=2024`
3 tháng 4 2023
=>\(\left(\dfrac{x-1}{2023}-1\right)+\left(\dfrac{x-2}{2022}-1\right)=\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-4}{2020}-1\right)\)
=>x-2024=0
=>x=2024
H
25 tháng 6 2023
\(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
Vì \(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
\(\Rightarrow\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}+4=0\)
\(\Rightarrow\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)+\left(\dfrac{x+21}{2023}+1\right)+\left(\dfrac{x+20}{2024}+1\right)=0\)
\(\Rightarrow\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}+\dfrac{x+2044}{2023}+\dfrac{x+2044}{2024}=0\)
\(\Rightarrow\left(x+2044\right)\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\right)=0\)
\(\Rightarrow x+2044=0\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\ne0\right)\)
\(\Rightarrow x=-2024\)
8 tháng 4 2022
refer
https://lazi.vn/edu/exercise/634984/tim-x-biet-x-1-2019-x-2-2020-x-3-2021x-4-2022
NH
22 tháng 2 2023
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
27 tháng 1 2022
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
27 tháng 3 2021
Câu 1:
1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{7}{x^2+3}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 4 2021
Câu 2a đề sai, pt này ko giải được
2b.
\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)
Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)
\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)
3a.
\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)
\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)
3b
Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6
\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)
Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
GH
29 tháng 6 2023
a
ĐK: \(x\ne5\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \Leftrightarrow\dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \Leftrightarrow\left(x-5\right)^2=12.3=36\\ \Leftrightarrow\left\{{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=11\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
b
ĐK: \(x\ne0;x\ne-1\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x}.\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2023}{2024}\\ \Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\\ \Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2023}{4048}=\dfrac{1}{4048}\\ \Leftrightarrow4048=x+1\\ \Leftrightarrow x=4047\left(tm\right)\)
29 tháng 6 2023
a: =>(x-5)/3=12/(x-5)
=>(x-5)^2=36
=>x-5=6 hoặc x-5=-6
=>x=11 hoặc x=-1
b: =>\(2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2023}{2024}\)
=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=2023/4048
=>1/2-1/x+1=2023/4048
=>1/(x+1)=1/4048
=>x+1=4048
=>x=4047
\(\dfrac{x-4}{2020}+\dfrac{x-3}{2021}+\dfrac{x-2}{2022}+\dfrac{x-1}{2023}+\dfrac{x-2024}{5}=4\) (sửa đề)
\(\Rightarrow\left(\dfrac{x-4}{2020}-1\right)+\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-2}{2022}-1\right)+\left(\dfrac{x-1}{2023}-1\right)+\dfrac{x-2024}{5}=0\)
\(\Rightarrow\dfrac{x-2024}{2020}+\dfrac{x-2024}{2021}+\dfrac{x-2024}{2022}+\dfrac{x-2024}{2023}+\dfrac{x-2024}{5}=0\)
\(\Rightarrow\left(x-2024\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}\right)=0\)
\(\Rightarrow x-2024=0\) (vì \(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}\ne0\))
\(\Rightarrow x=2024\)
\(\dfrac{x-4}{2020}-1+\dfrac{x-3}{2021}-1+\dfrac{x-2}{2022}-1+\dfrac{x-1}{2023}-1+\dfrac{x-2024}{5}+2=0\)
\(\Leftrightarrow\dfrac{x-2024}{2020}+\dfrac{x-2024}{2021}+\dfrac{x-2024}{2022}+\dfrac{x-2024}{2023}+\dfrac{x-2024}{5}+2=0\)
\(\Leftrightarrow\left(x-2024\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}\right)+2=0\)
\(\Leftrightarrow x=-\dfrac{2}{\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}}+2024\)