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Phuong Tran

CMR với mọi số thực x ta luôn có

\(\left[3x\right]=\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]\)

Nguyễn Việt Lâm
3 tháng 5 2019 lúc 23:10

Đặt \(x=\left[x\right]+\left\{x\right\}\)

\(\Rightarrow\left[3x\right]=\left[3\left[x\right]+3\left\{x\right\}\right]=3\left[x\right]+\left[3\left\{x\right\}\right]\)

\(\left[x+\frac{2}{3}\right]=\left[\left[x\right]+\left\{x\right\}+\frac{2}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]\)

\(\left[x+\frac{1}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

\(\Rightarrow\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]=3\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

Ta cần chứng minh \(\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

- Nếu \(\frac{2}{3}\le\left\{x\right\}< 1\Rightarrow\left\{{}\begin{matrix}2\le\left[3\left\{x\right\}\right]< 3\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\1\le\left[\left\{x\right\}+\frac{1}{3}\right]< 2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=2\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=1\end{matrix}\right.\)

\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

- Nếu \(\frac{1}{3}\le\left\{x\right\}< \frac{2}{3}\Rightarrow\left\{{}\begin{matrix}1\le\left[3\left\{x\right\}\right]< 2\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\0\le\left[\left\{x\right\}+\frac{1}{3}\right]< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=1\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\)

\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

- Nếu \(0< \left\{x\right\}< \frac{1}{3}\) tương tự trên ta có:

\(\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=0\\\left[\left\{x\right\}+\frac{2}{3}\right]=0\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\) \(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)


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