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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
5x5 +5y5 +2012z2012 =2022
Suy ra: 5x5 +5y5 + 2010z2012 +2z2012=2022
5(x5 +y5 + 402z2012 )+2z2012 =2022
Mà x,y,z là các số nguyên dương cho nên 5(x5 +y5 + 402z2012 ) là số nguyên dương
Cho nên 2022> 2z2012
Suy ra z chỉ có thể bằng 1 vì z là số nguyên dương
Thay z =1 vào biểu thức 5(x5 +y5 + 402z2012 )+2z2012 =2022, ta có:
5(x5 +y5 + 402 )+2 =2022
5(x5 +y5 + 402 )=2020
x5 +y5 + 402=404
x5 +y5=2
mà x,y là số nguyên dương nên chỉ thõa mãn khi x=1, y=1
Vậy x=y=z=1
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\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-5\right)^{2020}+\left(y-x+1\right)^{2022}=0\left(1\right)\)
Ta có \(\left\{{}\begin{matrix}\left(x-5\right)^{2020}\ge0,\forall x\\\left(y-x+1\right)^{2022}\ge0,\forall x;y\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^{2020}=0\\\left(y-x+1\right)^{2022}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-5=0\\y-x+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\y-5+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)
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A(1/2^2022)=1/2^2022+1/2^4044+...+1/2^(2022^2021)
=>2^2022*A=1+1/2^2022+...+1/2^(2022^2020)
=>A*(2^2022-1)=1-1/2^(2022^2021)
=>\(A=\dfrac{2^{2022^{2021}}-1}{2^{2022}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa: \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\le0\)
Mà \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\ge0\) với mọi x,y
Do đó \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{18}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=6+18=24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
`(x-1)^{2020}=(x-1)^{2022}`
`<=>(x-1)^{2022}-(x-1)^{2020}=0`
`<=>(x-1)^{2020}[(x-1)^{2}-1]=0`
`@TH1: (x-1)^{2020}=0<=>x-1=0<=>x=1`
`@TH2: (x-1)^{2}-1=0<=>(x-1)^{2}=1`
`<=>x-1=1` hoặc `x-1=-1`
`<=>x=2` hoặc `x=0`
\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)
⇒\(\left(x-1\right)^{2020}-\left(x-1\right)^{2022}=0\)
⇒\(\left(x-1\right)^{2020}\cdot\left[1-\left(x-1\right)^2\right]=0\)
+) \(\left(x-1\right)^{2020}=0\)
\(\Leftrightarrow x=1\)
+) \(1-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2=1\)
\(\Leftrightarrow x-1=1hoặcx-1=-1\)
\(\Leftrightarrow x=2hoặcx=0\)