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Ta có: \(\left(x+2022\right)⋮\left(x+5\right)\)
\(\Leftrightarrow\left(x+5\right)+2017⋮\left(x+5\right)\)
\(\Leftrightarrow2017⋮\left(x+5\right)\)
Vì \(x\in Z\) nên \(\left(x+5\right)\inƯ\left(2017\right)=\left\{\pm1;\pm2017\right\}\)
Ta có bảng sau:
x+5 | 1 | -1 | 2017 | -2017 |
x | -4 | -6 | 2012 | -2022 |
Vậy \(x\in\left\{-4,-6,2012,-2022\right\}\)
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\(\left(x-6\right)^{2020}+2\left(y-3\right)^{2020}=0\)
Ta có : \(\left(x-6\right)^{2020}\ge0\forall x\)
\(2\left(y+3\right)^{2020}\ge0\forall y\)
=>\(\left(x-6\right)^{2020}+2\left(y+3\right)^{2020}\ge0\forall x,y\)
Dấu "=" xảy ra <=>\(\left\{{}\begin{matrix}x-6=0\\y+3=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)
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(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). 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
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\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)
=>1/x+1=-1009/2022
=>x+1=-2022/1009
hay x=-3031/1009
x + (x + 1) + (x + 2) + ... + (x + 2022) + 2022 = 2022
x + x + x + ... + x + 1 + 2 + 3 + ... + 2022 + 2022 = 2022 (1)
Số số hạng x:
2022 - 0 + 1 = 2023 (số)
Từ (1) ta có:
2023x + 2022.2023 : 2 + 2022 = 2022
2023x + 2045253 = 2022 - 2022
2023x = 0 - 2045253
2023x = -2045253
x = -2045253 : 2023
x = -1011
Ta có : x + (x + 1) + (x + 2) + ... + (x+2022) + 2022 = 2022
=> x + (x + 1) + (x + 2) + ... + (x + 2022) = 2022 - 2022
=> [x + (x + 2022) ] . { [ (x + 2022) - x) : 1 + 1] } : 2 = 0
( số đầu + số cuối . số số hạng : 2 )
=> (2x + 2022) . 2023 : 2 = 0
=> 2x + 2022 = 0 . 2 : 2023= 0
=> (2x + 2022) : 2 = 0 : 2
=> x + 1011 = 0 => x = -1011