Trước hết ta chứng minh đẳng thức sau:
Với mọi số thực x, ta luôn có:
\(\left[3x\right]=\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]\) (1)
Thật vậy, đặt \(x=\left[x\right]+\left\{x\right\}\)
\(\Rightarrow\left[3x\right]=\left[3\left[x\right]+3\left\{x\right\}\right]=3\left[x\right]+\left[3\left\{x\right\}\right]\)
\(\left[x+\frac{2}{3}\right]=\left[\left[x\right]+\left\{x\right\}+\frac{2}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]\)
\(\left[x+\frac{1}{3}\right]=\left[\left[x\right]+\left\{x\right\}+\frac{1}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
Thay vào (1) trở thành:
\(\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\) (2)
Vậy ta chỉ cần chứng minh (2)
- Nếu \(\left\{x\right\}\ge\frac{2}{3}\Rightarrow\left[3\left\{x\right\}\right]=2\) ; \(\left[\left\{x\right\}+\frac{2}{3}\right]=1\); \(\left[\left\{x\right\}+\frac{1}{3}\right]=1\)
\(\Rightarrow2=1+1\) (đúng)
- Nếu \(\left\{x\right\}< \frac{1}{3}\Rightarrow\left[3\left\{x\right\}\right]=0\); \(\left[\left\{x\right\}+\frac{2}{3}\right]=0\); \(\left[\left\{x\right\}+\frac{1}{3}\right]=0\)
\(\Rightarrow0=0+0\) (đúng)
- Nếu \(\frac{1}{3}\le\left\{x\right\}< \frac{2}{3}\Rightarrow\left[3\left\{x\right\}\right]=1\) ; \(\left[\left\{x\right\}+\frac{2}{3}\right]=1\); \(\left[\left\{x\right\}+\frac{1}{3}\right]=0\)
\(\Rightarrow1=0+1\) (đúng)
Vậy đẳng thức (1) được chứng minh xong
Phương trình đã cho trở thành:
\(2\left[3x\right]=\left[3x\right]+1\)
\(\Leftrightarrow\left[3x\right]=1\)
\(\Leftrightarrow\frac{1}{3}\le x< \frac{2}{3}\)