High School Math Elite Teacher, Please Come Here! Collection Question 20

1 Attention! A may be or empty set!

2 a is not advisable for any number that is not 1. Substituting a=1 3 into (1+a) (1-a) gives 2, so 2 a. Substituting 2 into (1+a) (1-a) gives -3, hence -3 a.

Substituting -3 into (1+a) (1-a) gives -1 2, so -1 2 a. Then substitute -1 2 into 1 3, and so on, until the result of a is 1 and the other elements in the set are 2, -3, -1 2.

This is a question in our exam, hehe, please give points...

It could be 1 or 2 or 3. From '', we can see that A is an element and not a set, so A may be.

Do not take any number that is not 1. Substituting a=1 3 into (1+a) (1-a) gives 2, so 2 a. Substituting 2 into (1+a) (1-a) gives -3, hence -3 a.

Substituting -3 into (1+a) (1-a) gives -1 2, so -1 2 a. Replace -1 2 with 1 3 and start the cycle. So the result should be 2, -3, -1 2.

1.The A you write illustrates that A is just an element and not a set. A can only take one value, and it is not possible to take multiple values at the same time.

So a = 1 or 2 or 3 three values.

2.In the first case, a = 1 3.

1+a>/<1-a>=2

2 a Second.

1+a>/<1-a>=1/3

a=-1/2

So the other elements in a are 2 and -1 2

1. Yes, but to write it in the form of a set, an empty set is also possible.

2, the meaning of the second question is that when a is equal to 1 3, the other elements in the set; or (1+a) (1-a) is equal to 1 3 when other elements.

It should be 2,.

1.Yes. is not an arbitrary number that is not 0, such as 1; The question does not explain which numbers A takes. Just say a is desirable 1 3

is the element that lets us find the set when a=1 3.

1。That's right, a is just an element, i.e. it can only be one in a set.

2。I think the meaning of the topic is vague, and it is obviously not the original intention to let a take any number that is not 0 or 1, whether it is more reasonable to ask for (1+a) (1-a)=1 3 to push out a=-1 2 or a=1 3 to find (1+a) (1-a).

The answer to the first question is:

Other elements 2 and -1 in the second questionIn the second question, I think a is preferable to any number that is not 1. The literal meaning is not the element when we ask us to find a=1 3.

a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a=, or a

Question 2. (1+a) (1-a)=1 3 Let's substitute it too. Maybe there's a trade-off.

Question 1 A may be 1, 2 or 3

Question 2 The meaning of the text is to let you ball the other elements in the set when a is 1 3.

The first question is yes.

The second one is so confused. Can you just post the question? ^_

1,c={2,3,5} True subset:空集,{2},{3},{5},{2,3},{2,5},{3,5}

2,a=,b= acrossb={y |y=1}3,a=empty set,b={x |x=1 a} If a intersects b=b, then b must be a subset of a, so b is also an empty set, then a=0

For the concept of true subset, take the set a={1,2,3} as an example, if set b has less than or equal to 3 elements in a, then b is called a subset of a (note that less than or equal to, the number of elements in b can be 0,1,2,3); If set b has less than 3 elements in a, then b is said to be a subset of a (note that the number of elements in b can be 0, 1, 2).

The process is as follows; c=So the true subset is 2 3 5 2,3 2,5 5,3

a={1,2,3,4,5,6},,b={prime number}, so the element of c is 2,3,5, according to the formula, the answer is 2 to the 3rd power minus 1=7

Find the number of subsets, subtract 1 from the true subset of 2 to the nth power and subtract 1 The empty set is the true subset of any set, and the equation x -2x + 1 = -x +1 is solved to get x

a b=b indicates that b is contained in a a={x | x²+3x+4=0=｛1，-4｝

a-1 = 0 or -4 a-1 = 0 a = 1 or a = -1 4

1.Simultaneous y=ax+1, y=|x|, get ax+1=|x|1) at x 0, ax+1=x, a≠1, x=1 (1-a)2) x 0, ax+1=-x, a≠-1, x=-1 (1+a) because a, b intersect as a single element, so when x 0, x=1 (1-a)<0 or x<0, x=-1 (1+a) 0

1/(1-a)<0,a>1

1/(1+a)≥0,a<-1

3) When a=1, x+1=|x|, x=-1 2, the condition is satisfied;

4) When a=-1, -x+1=|x|, x=1 2, the condition is satisfied, so the value range of a is a 1 or a -1

2.(a b) c = (a c) (b c)1) bibison ax + y = 1, x 2 + y 2 = 1

x^2+(1-ax)^2=1

a^2+1)x^2-2ax=0

x[(a^2+1)x-2a]=0

x1=0,x2=2a/(a^2+1)

2) Simultaneous x+ay=1, x 2+y 2=1

1-ay)^2+y^2=1

a^2+1)y^2-2ay=0

y[(a^2+1)y-2a]=0

y1=0,y2=2a/(a^2+1)

So (a b) c must pass the points (0,1) and (1,0) so that (a b) c has only two elements, then x2 = 2a (a 2+1) = 0 or 1, y2 = 2a (a 2 + 1) = 0 or 1

Get a = 0 or 1

To make (a b) c have three elements, then the point (2a (a 2+1), 2a (a 2+1)) is the third element, substituting it into the circular equation.

2a/(a^2+1)]^2+[2a/(a^2+1)]^2=18a^2=(a^2+1)^2

a^4-6a^2+1=0

a^2=(6±√32)/2=3±2√2

a=1 2, or a=-1 2

1.The set a can be seen as a straight line with the slope of the point (0,1) and the set b as the first.

The angular bisector of the first and second quadrants, as shown in the diagram: a>1 or a<-1, there is only one element in the set a and b.

2。The set c is a circle with the origin as the center and 1 as the radius; The set a is a straight line with a slope of the crossing point (1,0), the set b is a straight line symmetrical with respect to the straight line y=x with respect to the set a, and the passing point (1,0); Obviously, no matter what value a takes, (a and b) always contain two elements, (0,1) and (1,0).

1) (A and B) intersect C with only two elements, then the line represented by A and B and the circle represented by C should have only one intersection point, i.e., A=0;

2), (a and b) intersection c contains three elements, then in addition to (1,0) and (0,1) there is only one intersection point, that is, the line represented by a,b and the circle represented by c should have two intersection points respectively, and the third intersection point is the same point. Because the straight line represented by a and b is symmetrical with respect to the straight line y=x, the third intersection point is on the straight line y=x, that is, the point (the root of the negative half is two, the root of the negative half is two), and the coordinates of this point are substituted into a (b can also be used) to find a = 1 + root 2.

The whole idea of solving the problem depends on drawing a figure to get help!

Question 1:

From b we get y=x or y=-x

1) y=x.

Substituting a x=ax+1

x=1/(1-a)

So y=|1/(1-a)|

2) y=-x.

Substituting a x=ax+1

x=1/(1+a)

So y=|1/(1+a)|

Since a and b are a set of single elements, the results obtained by (1) and (2) should be equal, i.e.,

1/(1-a)＝1/(1+a)

Get a=0 The idea of the second question is that ab is two straight lines c is a circle to draw a straight line and a circle, because a is changing, so the number of intersection points is also changing, the number of intersection points is (a and b) the number of intersection c elements The landlord tried to do it I haven't done a high school problem for a long time

1.From the image (the image of the two equations has one and only one intersection point) a 1 or a -12When an image passes (0,1) and (1,0) points a = 0 or 1, (a and b) intersect c with two elements.

When the intersection of the lines is exactly on the circle, (a and b) and c have three elements that are inverse functions of each other, and with respect to y=x symmetry, the intersection points are ( 2 2, 2 2) or (- 2 2, - 2 2) substituting to obtain a = 2-1 or 2+1

1 The intersection of two sets is a single element set, then y=ax+1 and y= x have only one intersection point, with the idea of combining numbers and shapes, draw an image of y= x, and it can be obtained from image analysis that only when the slope of y=ax+1 is 1 or -1, the two images have only one intersection point, so a=1 or aha=-12

1.Solution: can be solved by drawing method, each set can obviously be regarded as a set of coordinates of points, that is, line segments, for a, it is obviously a straight line with an intersection of y axis size of 1, b is a straight line y = x when x > 0 in the coordinate system, x < 0 is a straight line y = -x, if a and b intersect as a single element set, then there are only two cases, that is, the straight line represented by a is parallel to the straight line represented by b, when parallel to y = x, a = 1, and when parallel to y = -x, a = -1.

2.Solution: (1) a is the same as a in 1 represents a straight line, and b can be converted into y = (-1 a)x + 1 a); c represents a circle with a radius of 1 passing through the point (0,0), and when a is what is the value, (a and b) intersect c with two elements, that is, when a line and a line b intersect exactly two points on the diagram that intersect the circle c, a is equal to how much.

You can refer to question 1 to draw **.

2) In the same way, when a is what is the value, (a and b) intersect c with three elements, that is, when the points of a line and a b line intersect the c circle on the graph add up to exactly two, a is equal to how much. You can draw ** out.

Graphing plays a very important role in mathematics, especially if you come up with this kind of problem, you must master it. Good luck with your studies!

In the first question, you can draw the graph of y=ax+1 and y=x, because a and b intersect as a single element set.

So there is only one intersection point between two straight lines.

And a is over the fixed point (0,1).

From the graph, it is found that the value range of a is [1, positive infinity], [negative infinity, -1] Question 2.

Again, it has to be combined with several lines.

c is a circle with a radius of 1 at the origin of the center.

a over the fixed point (0,1) on the circle.

b over the fixed point (1,0) on the circle.

1) A and b) intersection c has two elements, which means that there can only be two intersections between two straight lines and a circle.

It can only be (0,1) and (1,0).

So a and b are the same straight line, i.e. a=1

2), (a and b) intersection c has three elements, and drawing means that there can only be three intersections of two straight lines and circles.

In addition to (0,1) and (1,0), there is one that must be the intersection of a,b on the circle ax+y=1 (1).

x+ay=1 （2）

x^2+y^2=1 （3）

Simultaneous equations (1), (2), and (3) can be solved to a.

Hope you understand.

There is also a combination of several lines is a very important method, I hope you can use it well in the future

Both of these problems are typical of the combination of graphics and algebra, and if you don't understand it, you can communicate privately.

Problems can be plotted on the coordinate axis for a more intuitive display.

Set b y=|x|On the coordinate axis are two perpendicular straight lines y=x and y=-x

The set a y=ax+1 is a straight line with a slope a

Then the meaning of the intersection of these two sets as a single element set is:

The line of set A has only one intersection point with the line of set B.

If a line has only one intersection point with two other lines that are not parallel to each other, then the line (y=ax+1) must be parallel to one of the two lines (y=x and y=-x). (otherwise there are two intersections anyway).

So it's easy to get a=1 or a=-1

Question 2: Set A Set B is a straight line, and Set C is a circle.

Positional relationship between a line and a circle:

Intersect There are two intersection points.

Tangent has an intersection point.

There is no intersection point apart.

A and b) intersect C, and there are two elements, that is, two straight lines and circles A and B, and C, for a total of two intersections.

Two cases: Both straight lines are tangent to the circle; One line intersects the circle and the other separates.

The radius of the circle is 1

The distance between the circle and the line a is 1 (a 2+1) (1 2).

The distance between the circle and the line b is 1 (a 2+1) (1 2).

The distance formula from a point to a straight line should have been learned) These two distances are the same, then there is only one possible case, that is, both lines are tangent to the circle.

1/(a^2+1)^(1/2)=1 a=0

3 (2) As we have just analyzed, the distance between these two lines and the center of the circle is the same, so their positional relationship to the circle is the same. So how do you have 3 intersections? are all apart, 0 intersections; both tangent, two intersections; All intersect, four intersections.

If there are only 3 intersections, then the meaning is that both lines intersect the circle, and two of the intersections coincide, that is, the intersection of the two lines themselves is on the circle.

Separation condition: 1 (a 2+1) (1 2) 1 solves a≠1

The intersection of two straight lines is the equation that joins the two straight lines;

ax+y=1

x+ay=1

Intersection (1 (a+1), 1 (a+1)).

This intersection is on a circle, then 1 (a+1) 2+1 (a+1) 2=1

a=2 (1 2)-1 or a= -2 (1 2)-1