Note: "e" is belonging to; "^" is the square; "C" is included in the set; "/" in the set is the column in the set, not divided by; Given the set a = {X / ax + B = 1}, B = {X / ax + b > 4}, where a is not equal to 0, if the element in a must be the element in B, find the value range of real number B Solution: the element of a is x = (1-B) / A and E B, so a * (1-B) / a - b > 4 (in this step, I don't understand why it's - B, isn't it + B? Because ax + b > 4, x = (1-B) / a, so substituting in is not a * (1-B) / A + right? I don't understand.) 4. Because of the restriction of element C {1,5} in nonempty set s. I don't quite understand this. Can you explain it in detail? I understand everything else. Also, do you have any good reference books to recommend to me?

Note: "e" is belonging to; "^" is the square; "C" is included in the set; "/" in the set is the column in the set, not divided by; Given the set a = {X / ax + B = 1}, B = {X / ax + b > 4}, where a is not equal to 0, if the element in a must be the element in B, find the value range of real number B Solution: the element of a is x = (1-B) / A and E B, so a * (1-B) / a - b > 4 (in this step, I don't understand why it's - B, isn't it + B? Because ax + b > 4, x = (1-B) / a, so substituting in is not a * (1-B) / A + right? I don't understand.) 4. Because of the restriction of element C {1,5} in nonempty set s. I don't quite understand this. Can you explain it in detail? I understand everything else. Also, do you have any good reference books to recommend to me?

1. There is a problem in the first question: a = {X / ax + B = 1}, B = {X / ax + b > 4}. In any case, the elements in a can not be included in B. because 1 is less than 4. Maybe the title is wrong
2. X ^ 2 + ax + B = x, move to x ^ 2 + (A-1) x + B = 0
3. The title should be:
Let u = {1,2,3,4,5}, a = {X / x ^ 2-5x + a = 0}, and an empty set be a proper subset of a, and a be a proper subset of U. find the value of a and the complement of a in U
Because an empty set is a proper subset of a, a is not an empty set. And a is a proper subset of u, so a! = (not equal to) U. that is, x ^ 2-5x + a = 0 has a solution, and the solution is in U, so 5 ^ 2-4a > = 0, a ≤ 5 / 2, and each number of u is substituted into a = 4, a = 6, a = 4, a = 0, so a takes 0, and a = {X / 5}, so the complement of subset a in U: {1,2,3,4}
4. Because of the restriction of element C {1,2,3,4,5} in nonempty set s
5. Remove the absolute value
6. It's very easy to remove the absolute value. There are only two cases. One is that the absolute value is positive, and it's OK to remove the absolute value directly. The other is negative, and it's OK to add a negative sign
Finally, I would like to say that the reference book you sell is not good. Try other ones without mistakes and analysis of answers. Of course, you have to do them first. When you understand these, you can just look at the answers instead of doing them

1、 If the points a (2,2), B (a, 0), C (0, b), (AB ≠ 0) ABC are collinear, then one part of a + one part of B is equal to?
2、 Given SiNx = 2cosy, find the value of Sin & sup2; y + 1 / 2-sinycosy
3、 The maximum value of radical 2 × cos (2x + quarter π) on X ∈ [0, half π]
Change x to y

1、 If the points a (2,2), B (a, 0), C (0, b), (AB ≠ 0) ABC are collinear, then one part of a + one part of B is equal to?
A. The equation of the straight line where B and C are located can be written as intercept formula: X / A + Y / b = 1, a (2,2) is on the straight line
2 / A + 1 / b = 1, so 1 / A + 1 / b = 1 / 2
2、 Given siny = 2cosy, find the value of Sin & sup2; y + 1 / 2-sinycosy
∵siny=2cosy,∴tany=2,∴sin²y=4/5,cos²y=1/5
So (2-sinycosy) / (Sin & sup2; y + 1) = (2-2cos & sup2; y) / (Sin & sup2; y + 1) = 2 (1-1 / 5) / (4 / 5 + 1) = (8 / 5) / (9 / 5) = 8 / 9
3、 The maximum value of y = (√ 2) cos (2x + π / 4) on X ∈ [0, π / 2]
It is easy to determine that when x = 0, ymax = (√ 2) cos (π / 4) = (√ 2) (√ 2 / 2) = 1 by using the five point drawing method

It is known that the two focal points of the ellipse are on the coordinate axis and symmetric about the origin. The focal length is 6. The ellipse passes through the point (0,4). The standard equation of the ellipse is obtained

The focal length is 6, so C = 3, we can know that the focus should be on the X axis, so from the ellipse passing through the point (0,4), we know that B = 4, so a = 5, so the standard equation is
X square / 25 + y square / 16 = 1
The people upstairs said it was an ellipse