It is known that proposition p: there exists x belonging to R such that x ^ 2-2ax + 2A ^ 2-5A + 4 = 0; proposition q: the curve X ^ 2 / 3 + y ^ 2 / A-3 = 1 is hyperbola. If "P or Q" is true and "P and Q" is false, the value range of real number a is obtained

It is known that proposition p: there exists x belonging to R such that x ^ 2-2ax + 2A ^ 2-5A + 4 = 0; proposition q: the curve X ^ 2 / 3 + y ^ 2 / A-3 = 1 is hyperbola. If "P or Q" is true and "P and Q" is false, the value range of real number a is obtained

Two propositions are given: proposition a: the solution set of the inequality x2 + (A-1) x + A2 ≤ 0 about X is ∈; proposition B: the function y = (2a2-a) x is an increasing function. (1) at least one proposition of a and B is true; (2) there is and only one proposition of a and B is true; the value range of real number a in accordance with (1) (2) is obtained respectively

When a is a true proposition, △ = (A-1) 2-4a2 ＜ 0, the solution is a ＞ 13 or a ＜ 12, that is, when a = {a | a ＞ 13 or a ＜ 12} B is a true proposition, 2a2-a ＞ 1, the solution is a ＞ 1 or a ＜ 12, that is, B = {a | a ＞ 1 or a ＜}

When a and B satisfy what conditions, the equation 3ax = x + B (1) has unique solution (2) has innumerable solutions (3) has no solution

Solution
3ax=x+b
（3a-1)x=b
(1) The equation has a unique solution, that is 3a-1 ≠ 0, so when a ≠ 1 / 3, the equation has a unique solution
(2) The equation has no solution, that is 3a-1 = 0, B ≠ 0, so when a = 1 / 3, B ≠ 0, the equation has no solution;
(3) The equation has innumerable solutions, that is 3a-1 = 0, B = 0, so when a = 1 / 3, B = 3, the equation has innumerable solutions

It is known that the equation 2x + 6A = 3ax - (X-12) ① with respect to X has no solution, then a = ② with innumerable solutions, then a = ③ with unique solution, then a=

When there is no solution, (3a-3) x = 6A + 12, if (3a-3) = 0 and 6A + 12 does not = 0, the solution is a = 1;
When there are innumerable solutions, if 6A + 12 = 0 and (3a-3) does not = 0, the solution is a = - 2; when there is a unique solution, that is, 6a + 12 does not = 0 and (3a-3) does not = 0, the solution is a does not = 1 and a does not = - 2