The problem of the collection, I hope that the master will solve the doubt, and some questions about

m<=-1

The process of solving the problem is:

First of all, there are three cases:

1): Equation x2-4mx+2m+6=0, there are 2 different negative roots, then you can get a system of equations.

16m2-4 (2m+6) >0 (discriminant greater than zero), 4m<0 (the sum of the two is less than zero), 2m+6>0 (the product of the two is greater than zero), 2m<0 (the axis of symmetry is on the left side of the y-axis).

In the first case, these four equations give a range of m: m<-1

2): The equation has only one negative root, so these are the following equations.

16m2-4(2m+6)>0 (discriminant greater than zero), 2m+6<=0 (the product of two is less than or equal to zero).

The range of m obtained: m<=-3 2

3): The equation has two identical negative roots, and the system of equations is:

16m2-4(2m+6) = 0 (discriminant equal to zero), 2m<0 (axis of symmetry to the left of the y-axis).

The range of m obtained: m=-1

Then find the union in the above three cases, and you can get the specific correct range of m: m<=-1, that is, m is less than or equal to minus one.

Explain that the equation x2-4mx+2m+6=0 has at least one negative root.

It can be divided into the following two situations:

1 A negative root and a non-negative root, then.

Discriminant formula 0, when x=0, the value of the function is less than 0

Discriminant = 0, axis of symmetry less than 0

2 Two negative roots, then.

Discriminant formula "0", two roots and < 0, two roots "0", can be found by using Veda's theorem.

I'm not going to do the math, you just do the algebra as described above.

x2-4mx+2m+6=0 has at least one negative root.

-4m)²-4(2m+6)=16m²-8m-24≥0 ①4m-√△/2<0 ②

Solution , 16m -8m -24 0

2m²-m-3≥0

2m-3)(m+1)≥0

m -1 or m 3 2

Solution, (4m- 2<0

4m-√(16m²-8m-24)]/2<02m-√(4m²-2m-6)<0

2m<√(4m²-2m-6)

m<0 and 4m -2m-6 0 or m 0 and 4m -2m-6 0 and (2m) <4m -2m-6).

m<0 and 4m -2m-6 0 or m 0 and 4m -2m-6 0 and m<-3

m<0 and 4m -2m-6 0 or no solution.

So m<0 and 4m -2m-6 0

4m-2m-6 0 has been solved, which is m-1 or m3-2 so m-1

The answer is easy to know from the question x y=(x-y) (y-x)x y-x y

The following **x y, that is, the common solution of y=sinx and y=x2, the image is easy to know if there is an intersection point, then it must be in [0, 2].

Examine the function f(x)=x2-sinx, x [0, 2] then f'(x)=2x-cosx

f''(x)=2+sinx>0 is always established.

f'(x)min=f'(0)=2-1=1>0f(x) is also monotonically increasing, and f(x)min=f(0)=0 has only one common solution, which is x=0

a△b=

Discussion: 1, when a is an empty set, that is, 2a+1>3a-5, a<62, when a is not empty, that is, a>=6, there should be: 2a+1>=3 and 3a-5<=22, get: 1<=a<=9, so, 6<=a<=9

To sum up, take the union of 1 and 2 to get a<=9

This is why the union is finally taken because each case satisfies the question requirements.

The result is that 2a+1"3 hands in 3a-5"22, i.e., 1"a"9, it cannot be combined.

cua∩cub=cu（a∪b）=

a b=a cub is a has b no yes.

a b - (a cub -cu(a b)) is b, there is a, there is none, there is rest.

So a=b=check it yourself.。。

Shift to solve, is b={x|-2cra=｛x|x -1 or x 3}, so, cra) b = {x|-2 x -1 or x=3}

a=b=

You can push according to the topic.

What is the CRA in the second question?

Let m=, n=

The square term is constant and non-negative, x 0, x +1 1

m=x takes any real number, and y=x +1 is always meaningful.

n = r vs. r, the two are very different.

The two collections are not the same collection.