Find the range of xxx for which x2−4x+3x2+1≤0\frac{x^2 - 4x + 3}{x^2 + 1} \leq 0x2+1x2−4x+3≤0.
(1,3)(1, 3)(1,3)
(−∞,1]∪[3,∞)(-\infty, 1] \cup [3, \infty)(−∞,1]∪[3,∞)
[1,2][1, 2][1,2]
[1,3][1, 3][1,3]