Senior 1 set questions 20, senior 1 set questions.

Solution: x 2-3ax+2a 2=0

x-a)(x-2a)=0

x = a or 2a

1)a∩b=b

So when a>0, when a=0, when a<0.

At 00 hours, there is A&4

To sum up, in order for a b = empty set to hold, a can be valued in the range of (-infinity, -2] and [4,+infinity).

1) Solved by x 2-3ax + 2a 2 = 0, x = 2a, or = a;

Since a b = b, the set of values of a = {a|-2<2a<4 2=4 and 2a>=4, i.e., a>=4;

or a<=-2 and 2a<=-2, i.e., a<=-2

The answer on the 2nd floor is wrong, -2 and 4 must be taken to close the interval, that is, (-infinity, -2] and go up to [4, + infinity).

The easiest and clearest way:

Draw a quadratic function y=x 2-3ax+2a 2, and the abscissa of the intersection point with the x-axis is a,2a

1) A b=b, which means that b is within the range of a, that is, the abscissa of the two intersections is between (-2,4), then the value range of a is (1,2).

2) a b = empty set, indicating that the abscissa of the two intersections is outside (-2,4), then a>2 or a<-2

LZ, choose B, right?

p is an even set and q is an odd set.

a+b=even+odd=odd So it belongs to Q 978, I hope it helps you!

I think the landlord is wrong, "where ak is the remainder of i+j divided by 4" should be changed to "where k is the remainder of i+j divided by 4"? If yes, take a look below:

i, j are the subscripts of ai and aj, i, j = 0, 1, 2, 3 means that the values of i and j can be 0, 1, 2, 3 respectively

i.e. i 0 or 1 or 2 or 3; j 0 or 1 or 2 or 3;

Ideas for solving the problem. x@x) @a2=ao, compared to ai@aj=ak, you can think of (x@x) as ai, a2 as aj, where j 2, and a0 as ak, where k 0From the fact that 'where k is the remainder of i + j divided by 4', we know that i + 2 (note that j has been replaced by 2) has a remainder divided by 4 as 0, because i o or 1 or 2 or 3, and to make "i + 2 divided by 4 the remainder of 4 0", i can only be 2

That is, (x@x) A2, in the same way, X is regarded as AI and AJ in "ai@aj=AK", but i=

At this time, i+j; (because i=j), to make the remainder of i+j divided by 4 be k 2, i+j can only take 2, or 6i.e. i.e. i=j=1 or 3That is, x a1 or a3, so the number of x (x belongs to s) satisfying the relation (x@x) @a2 = ao is 2, that is, a1 or a3.

I don't know if the landlord can understand it, the level of contemptible explanation is limited, but the method is like this.

If the landlord has any questions, you can hi me.

a∩b=｛1/2｝

Then x=1 2 is the common root of the two equations.

Substitute x=1 and 2 respectively.

1/2-p/2+q=0 (1)

3 2+(p+2) 2+5+q=0 (2) so p=-7, q=-4

So a is 2x +7x-4=0

x+4)(2x-1)=0

x=4,x=1/2

b is 6x -5x + 1 = 0

3x-1)(2x-1)=0

x=1/3,x=1/2

So a b=

Because b=

So b=[1,2].

Scenario 1:a is an empty set.

then delta = 4-4a<0

So a>1

Scenario 2:a is not an empty set.

Because the parabola y=x 2-2x+a is x=1, once the parabola intersects the x-axis, two different real roots will be smaller.

Beyond the scope of b, because a is really included in b, it is incompatible.

So delta=4-4a=0

A=1 In summary, a>=1

It is easy to obtain the range of b set x is [1,2];

If a is not an empty set, and a is a true subset of b. Then the two roots of the equation x 2-2x + a=0 1<=x1<=2 1<=x2<=2

So 2<=x1+x2<=4; and x1+x2=2;Therefore it is only possible that x1=x2=1;a=1；

If a is an empty set, the condition is also satisfied. In this case, the discriminant formula is 4-4a<0; a>1;

Therefore, a>=1;

by b={x|3-x0}=, and a is really contained in b, then.

a=｛y|y=x 2+2x+a, and 34-(x+1) 2

Get a>4

cua=

cua)∪b=

cua)∪b]∩z=

So c= or c= or c=

c b ≠ empty set, so c= or c=

cua=

cua）∪b=

cua）∪b）∩z==

c is contained in ((cua) b) z

The element in c is an integer.

There is only -2 in b and -2 in c.

c is or.

I choose 8First of all, there is a default: a, b are both in u.

This should be common knowledge for this type of question. From condition 8, we can see that there are no 8 or 8 in A and B (Morgan's rule is used here, which your teacher should have talked about), and now we know that there can only be 8 or 8 or 8 or 8 in A and B. Using condition 8 and combining condition 8, the element in b is known to be 8,8. Hypothesis.

d consists of the uniqueness of the elements of the collection, and the elements inside cannot be the same.

So there can't be two sides equal, so it's impossible to form an isosceles triangle.

For a, y=x +2x+a=(x+1) +a-1 a-1 i.e. y a-1

So a={y|y a-1}, again a true contained in b for b, x 3

So A-1 3

So a 4

I don't remember the formula, but I can tell you how to solve it.

First solve b, know the range of the value of x, and a is really included in b, which means that the value of the set y should be greater than 3 when x is greater than or equal to 3, column function. And the function increases monotonically in the range of x greater than or equal to 3, indicating that the axis of symmetry should be less than or equal to 3. column inequalities, you should be able to solve the range of a.

Solution: y=x 2+2x+a is a parabola with the opening facing up, the minimum value is a-1, and a b, b==, so.

a-1 3 Therefore, a 4 is the range of values of a.