When the value of real number a is (a-3a + 2) x + (A-1) x + 2 > 0, it is always true for any X

The solutions of a ^ 2-3a + 2 = 0 are a = 1 and a = 2
When a = 1, the inequality is reduced to 2 > 0 for any x belonging to R
When a = 2, the inequality is reduced to x + 2 > 0, x > - 2
When a ≠ 1,2, the left side of the inequality is a quadratic function
In order to make the inequality hold, the parabolic opening of quadratic function should be upward, and there is no intersection with X axis (Δ 0, the solution is a > 2 or a)

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