Nếu x < - 2
=> |x - 1| = -(x - 1) = -x + 1;
|x + 2| = -(x + 2) = -x - 2;
|x - 3| = -(x - 3) = - x + 3
Khi đó |x - 1| + |x + 2| + |x - 3| = 14
<=> - x + 1 - x - 2 - x + 3 = 14
<=> - 3x + 2 = 14
=> x = -4 (tm)
Nếu -2 \(\le\)x\(\le\)1
=> |x - 1| = -(x - 1) = -x + 1;
|x + 2| = x + 2;
|x - 3| = -(x - 3) = - x + 3
Khi đó |x - 1| + |x + 2| + |x - 3| = 14
<=> -x + 1 + x + 2 - x + 3 = 14
=> - x = 8
=> x = - 8 (loại)
Nếu 1 < x < 3
=> |x - 1| = x - 1
|x + 2| = x + 2;
|x - 3| = -(x - 3) = - x + 3
Khi đó : |x - 1| + |x + 2| + |x - 3| = 14
<=> x - 1 + x + 2 - x + 3 = 14
=> x = 10 (loại)
Nếu x > 3
=> |x - 1| = x - 1
|x + 2| = x + 2;
|x - 3| = x - 3
Khi đó : |x - 1| + |x + 2| + |x - 3| = 14
=> x - 1 + x + 2 + x - 3 = 14
=> 3x = 16
=> x = 16/3 (tm)
Vậy \(x\in\left\{\frac{16}{3};-4\right\}\)
