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\(A=8\left(1+8\right)+8^3\left(1+8\right)+...+8^{2021}\left(1+8\right)\)
\(=8.9+8^3.9+...+8^{2021}.9=9\left(8+8^3+...+8^{2021}\right)⋮9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(y^2=3-\left|2x-3\right|\)
Vì \(-\left|2x-3\right|\le0\forall x\) nên \(3-\left|2x-3\right|\le3\forall x\) nên \(y^2\le3\rightarrow y^2\in\left\{0;1\right\}\) (vì \(y\in Z\) )
TH1:
\(y^2=0\)
\(\Rightarrow y=0\)
\(\Rightarrow\left|2x-3\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
TH2:
\(y^2=1\)
\(\Rightarrow y=\pm1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) \(M=\frac{2019}{2020}+\frac{2020}{2021}\) và \(N=\frac{2019+2020}{2020+2021}\)
Ta có \(\frac{2019}{2020}>\frac{2019}{2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2020+2021}\)
\(\Rightarrow\frac{2019}{2020}+\frac{2020}{2021}< \frac{2019+2020}{2020+2021}=N\)
\(\Rightarrow M>N\)
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\(\Rightarrow8P=8^2+8^3+...+8^{2022}\)
\(\Rightarrow8P-P=8^2+8^3+...+8^{2022}-8-8^2-...-8^{2021}\)
\(\Rightarrow7P=8^{2022}-8\)
\(7P+8=8^y\\ \Rightarrow8^{2022}-8+8=8^y\\ \Rightarrow8^{2022}=8^y\\ \Rightarrow y=2022\)
em cảm ơn ạ