The math problem of set in four high schools is 100.

A∈A is -4

X+a= 1→x= 1-a∈(0，5)；

Then b = {x │ 0

Then a ∩ b = {x │ 0

A∩B={- 1}, with:

1+p+q = 0；

1-p-2q = 0；

To solve this system of equations:

p =-3； q=2。

Namely: A={x│x? -3x+2=o}，B={x│x？ +3x-4=0}

Solve the equation by decomposition:

A={ 1，2}，B={ 1，-4 }；

Then A∪B={ 1, 2, -4}

three

The key to solve this problem is to understand the meaning of the problem, UA = {5}, which means 5∈U, but 5A, so A2+2A-3 = 5.

| 2a- 1 | ≠ 5 and | 2a- 1 | ∈ u

Solution: ua = {5}, ∴5∈U and 5a.

∴ A2+2A-3 = 5, and the solution is A = 2 or A =-4.

When a = 2, | 2a- 1 | = 3 ≠ 5,

When a =-4, | 2a- 1 | = 9 ≠ 5 but 9U

∴ A =-4 (shed) ∴ A = 2.

four

There is an error in the question: x? +mx+n=0 is a quadratic equation, so there can only be two elements in a at most;

So the complement set of A in U should have at least two elements.