\(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=x\)=> x nguyên => x có thể có các dạng sau: 6k ; 6k + 1; 6k + 2; 6k +3 ; 6k + 4; 6k + 5 ( k nguyên)
+) Nếu x = 6k
PT <=> \(\left[\frac{6k}{2}\right]+\left[\frac{6k}{3}\right]=6k\) => \(\left[3k\right]+\left[2k\right]=6k\) => 3k + 2k = 6k => 5k = 6k => k = 0 => x = 0
+) Nếu x = 6k + 1
PT <=> \(\left[3k+0,5\right]+\left[2k+\frac{1}{3}\right]=6k+1\)<=> 3k + 2k = 6k + 1 <=> k = - 1 => x = -5
+) Nếu x = 6k + 2
PT <=> \(\left[3k+1\right]+\left[2k+\frac{2}{3}\right]=6k+2\) <=> 3k + 1 + 2k = 6k + 2 <=> k = -1 => x= -4
+) Nếu x = 6k + 3
PT <=> \(\left[3k+1,5\right]+\left[2k+1\right]=6k+3\) <=> 3k + 1 + 2k + 1 = 6k + 3 <=> k = -1 => x = -3
+) Nếu x = 6k + 4
PT <=> \(\left[3k+2\right]+\left[2k+\frac{4}{3}\right]=6k+4\) <=> 3k + 2 + 2k + 1 = 6k + 4 <=> k = -1 => x = -2
+) Nếu x = 6k + 5
PT <=> \(\left[3k+2,5\right]+\left[2k+\frac{5}{3}\right]=6k+5\) <=> 3k + 2 + 2k + 1 = 6k + 5 <=> k = -2 => x = -7
Vậy x \(\in\) {0; -5;-4;-3;-2;-7}
