What are the two math problems senior 1 one .

1.First a is a non-empty set, then there are 3a-5>2a+1, we get a>6 and then a is a subset of b, then 2a+1 3 and 3a-5 22, we get 1 a 9 and there is 62First, if m=0 and q are empty sets, they must be a subset of p.

If m≠0, then q=

The discriminant formula of the equation in p is equal to 25m and must be greater than 0 (m≠0), then the root finding formula p is.

If q is a subset of p, then there is 1 m = 3m or 1 m = -2m to get m = 3 3 In summary, the set of values of m is .

Because a is a non-null set.

So 3a-5 2a+1 is a 6

Since a is included in (a b).

So the intersection of A and B must contain all the elements in A.

So a is a subset of b [so that a b is a and a is contained in a] and thus it is obtained. It is 2a+1 3,3a-5 22

Solution 1 a 9

In summary, a 6,1 a 9

So the set of all a values that make a (a b) hold is.

The first one someone answered, so I didn't answer.

The second one: first calculate the value of p set x, use the matching method to get the 2nd power of (x-m 2) = (5m 2) to the 2nd power, turn it into x=3m or x= -2m, and then calculate the value of q set x as x=1 m, because q is a subset of p, so x in q must be one in p, that is, 1 m=3m or 1 m= - 2m, and get m=1 3 under the positive and negative root numbers, there should be no mistake, I haven't done it for a long time.

As for learning mathematics, I think I still have to understand the principles of those theorems! It's not that you know it's a theorem or an axiom, you have to know how it came about, you have to know how to prove it! This kind of math will be very good, hehe! Hope you understand.

1 2a+1 x 3a-5 is a subset of 3 x 22 which is 3 2a+1 3a-5 22 and calculates the range of a 1 a 9

1.Let 2 x t 0 x 2 1 t 4 then y t 2 2at 5

The quadratic function with respect to t is obtained and the axis of symmetry is t a 2

When a 2 4 i.e. a 8, t 4 at y, minimum 21 8a, when a 2 1 i.e. a 2, at t 1, y minimum 6 2a, when 1 a 2 4, i.e. 2 a 8, at t a 2, at 2 y, at least 5 (3 4) a 2

In summary, when a 8 x 2 y, a minimum 21 8a, when a 2 x 0, y a minimum 6 2a

When 2 a 8 x (a 2) (base 2 a 2 logarithm) 2When the definition domain of f(x) is (-.

ax 2-2x+2 0 to x r constant formation.

then a 0 and the minimum value (8a-4) 4a 0 is solved to get a 1 2

When the domain of the function f(x) is t and p t≠ i.e. ax 2-2x+2 0 has a solution on p.

Let t(x) ax 2-2x+2 ax axis of symmetry be x 1 a, and when a 0 is obviously not true.

When a 0 requires 2x 2 0 x 1 to have a solution on p, and when a 0 then 1 a 2 i.e. 0 a 1 2 needs t(1 2) 0 to solve a 4, then only 0 a 1 2

If 1 a 1 2 i.e. a 2 needs t(2) 0 to be solved a 1 2 rounded.

If 1 2 1 a 2 i.e. 1 2 a 2 requires t(1 2) 0 or t(2) 0

Solve 1 2 a 2

In summary, we get 0 A 2

1.The square satisfies all the conditions of a rhombus, so it can be said that b is a subset of a2Because there is only one and only one element in a, then.

a=0.or 2 2-4*a=0

There is a=0,1

When a=0, a=

When a=1, a=

1. Yes, it is known by the definition of the subset.

2. When a=0, x=, a=

When a≠0, the discriminant formula = 0, that is: 4-4a = 0, a = 1a

1.Since the squares are all diamond-shaped, b belongs to a, i.e. b can be said to be a subset of a.

There is only one element in select x and there is only one solution.

When a=0, x=

When a is not equal to 0. =2 -4a=0 and a=1 then there is a {0,1}

c2=a2+b2-2abcosc So according to the known conditions, c2=4+8-8 2cos15=12-2 2-4 3

c sinc=2 sina is too troublesome to fight against yourself and is too troublesome sinc=(root number 6 - root number 4) 4 The three conditions used to calculate can be calculated by sedan and spike and then calculate a

1. (1): -x belongs to (0, positive infinity).

f(-x)=-x^2-mx-1

Since it is an odd function, satisfying: f(-x)=-f(x)so, -f(x)=-x 2-mx-1

So, when x(0, positive infinity), f(x)=x2+mx+1(2): the odd function is symmetric with respect to the x-axis, b 2-4ac>0 -20 x>0 so the whole equation is greater than 0, monotonically increasing.

3) Since it is a monotonic increasing function, g(-2)=f(2)=5 2g(1 2)=f(-1 2)=-5 2

The value range is (-5 2, 5 2).

Solution: (1) Let x 0, then -x 0, f(-x)=-x2-mx-1(2 points).

f(x) is an odd function, i.e., f(-x) = -f(x), (3 points).

So, f(x)=x2+mx+1(x 0), (4 points).

f(0)=0, (6 points).

So f(x)=

x2+mx+1 x＞00 x=0-x2+mx-1 x＜0

7 points) 2) Because f(x) is an odd function, the image of the function y=f(x) is symmetrical with respect to the origin, (8 points).

From the solution of five unequal real numbers in the equation f(x)=0, the image of y=f(x) has five different intersections with the x-axis, (9 points).

f(0)=0, so the image of f(x)=x2+mx+1(x 0) has two different intersection points with the positive half axis of the x-axis, (10 points).

That is, the equation x2 + mx + 1 = 0 has two unequal positive roots, and the two roots are x1 and x2 (11 points) respectively

=m2-4＞0x1+x2=-m＞0x1•x2=1＞0

m -2, (14 points).

Therefore, the value range of the real number m is m -2 (15 points).

The question is wrong, how can a quadratic curve have 5 unequal real solutions?

In fact, what do you think is the use of making these problems, do you feel complacent when you see that others can't do it, or do you think that these so-called functions will help you face the cruel Chinese society in the future?

1. Let f(x)=kx 2x k

If k=0, then there is no unique solution;

If k≠0, then f(x) is a quadratic function, and to make a unique solution, then:

The minimum value of the opening is up and f(x) is 2 or the maximum value of the opening is down and f(x) is 1, only then can there be a unique solution;

2. According to the provisions of *, there are: (x y) * (x y) = (x y) (1 x y) = x x y y<1 i.e

With regard to the inequality of x: x x y y 1>0 to one real number x is constant, then only its discriminant is greater than 0.

Discriminant = ( 1) 4 1 ( y y 1)<0 to find the range of y.

kx^2+2x+k-1>=0

b^-4ac>=0

4-4k(k-1)>=0

Root number 2< = k< = root number 2

kx^2+2x+k-2<=0

b 2-4ac>=0 (with solid roots).

Root number 3< = k< = root number 3

So k [root number 2, - root number 2].

1.Calculate the solution set of kx 2+2x+k<1 kx 2+2x+k>2, and then find its complement.

2.（x-y）*（x+y）=（x-y）（1-x-y）=x-y-x^2+y^2

y^2-y+1/4-（x^2-x+1/4）=(y-1/2)^2-(x-1/2)^2

1So(y-1 2) 2<1+(x-1 2) 2<11 2-1