Set a = {x | - 1 less than or equal to x less than 2}, B = {x | x less than a}, if a intersection B is not equal to the empty set, find the value range of A Why can't a ＜ 2? It's also in that range, and there is intersection

Set a = {x | - 1 less than or equal to x less than 2}, B = {x | x less than a}, if a intersection B is not equal to the empty set, find the value range of A Why can't a ＜ 2? It's also in that range, and there is intersection

Set a = {x | - 1

The solution set of the inequality ax with square plus BX plus C greater than zero is the set (x | x is less than minus 2 or greater than minus 2 / 2) to find the square of ax minus BX plus C greater than zero

Drawing the image of quadratic function, we can know that the intersection of image and X axis is x = - 2, x = - 1 / 2
The expression of symmetry axis of quadratic function is: - B / 2A at the same time
Then the sign of the axis of symmetry of the two functions is changed, so the image is symmetric to the positive half axis of X
The two intersections become: x = 2, x = 1 / 2
Therefore, it is easy to get the answer: {x | x2}

The solution set of X inequality 1 plus BX is a if a is equal to 1b and negative 1

Is to solve the inequality x ^ 2 / (1-x) ≤ X
When 1-x > 0, x ^ 2 ≤ x (1-x)
The solution is 0 ≤ x ≤ 1 / 2
When 1-x1
So the solution set of inequality x ^ 2 / (1-x) ≤ x is {x | 0 ≤ x ≤ 1 / 2 or x > 1}

It is known that one of the univariate quadratic equation AX ^ 2 + BX + C = 0 (a is not equal to 0) is followed by 1, and a and B satisfy B = (root A-2) + (root 2-A) - 3. Find another root of the univariate quadratic equation

One of the univariate quadratic equations ax ^ 2 + BX + C = 0 (a is not equal to 0) of X is followed by 1
a+b+c=0.(1)
a. B meet the following requirements:
B = radical (A-2) + radical (2-A) - 3. (2)
The condition that the root sign (A-2) is meaningful is as follows
a≥2.（3）
The conditions for the root sign (2-A) to be meaningful are as follows
a≤2.（4）
From (3) and (4), it is concluded that:
a=2.(5)
Substituting (5) into (3) yields:
b=-3
Therefore, the original equation can be reduced to:
2x^2-3x+c=0
x^2-1.5z+0.5c=0
According to Weida's theorem:
x1+x2=1.5
x2=1.5-x1=0.5