Is this simple high school math problem right?

That's right, the range of the value range is r, that is, the value of the defined field meets all the values on f (x) and r, and the image can be obtained that u (x) and x axis must have an intersection. Hence >=0

If you define the domain as r, then <0, you can draw an image of the quadratic function to help with the analysis.

The question is correct, it's just that you don't grasp the points of investigation of this kind of topic!

The question condition gives a value range, whereas you usually encounter a definition domain.

The problem gives that the range of the value range is r, which means that the value of the defined domain meets all the values on r if f(x) can be removed, and the drawing shows that u(x) can take a value at any point in (0, infinity), and in order for u(x) to take any value, it can be obtained through the image that u(x) and x-axis must have an intersection point. Hence >=0

Similar to what you encounter in the Constant Formation.

That's right. The logarithmic function range is a real number, which means that all positive numbers can be obtained in the definition field (of course, some of the definition fields may not satisfy the natural definition domain, so these unsatisfactory ones will be automatically removed and do not need to be considered).

Naturally defined domains are self-satisfying, and can be ignored because there is no question about defining domains in the question.

The answer is right! Learning math well mainly makes me good concepts! Pay attention to the relationship between constant establishment and constant existence!

This is a high school difficulty and it is also a key point, there are two kinds, one is to define the domain as r, as long as the definition domain is meaningful.

The other is that the value range is r, that is, to ensure that all the books in the defined field are finished, for this problem. Your approach is right》0

If you change the condition to define the domain to r, then, <0

No, the title doesn't say that the definition domain is r, as long as f(x) is satisfied to get r, don't worry about defining the domain.

p；1 (x -x-2) >0, not p:1 (x -x-2) >0 does not hold.

Non-p is equivalent to: 1 (x -x-2) 0, or x -x-2 = 0, so your solution (1) is correct, and solution (2) is problematic.

Let a (1 3) = x, b (1 3) = y Simplify a-3a (2 3)+5a (1 3)=1x 3-3x 2+5x-1=0

x(x-1)(x-2)+3(x-1)=-2(x-1)((x-1)^2+2)=-2

x-1)^3+2(x-1)=-2

b-3b^(2/3)+5b^(1/3)=5y^3-3y^2+5y-5=0

y(y-1)(y-2)+3(y-1)=2

y-1)^3+2(y-1)=2

Function f(x')=x'^3+2x'=x'(x'^2+2)f(-x')=-f(x')

f(x') is an odd function with origin symmetry.

The above system of equations can be seen as f(x')=2 f(x')=-2, i.e. y-1=x0

x-1=-x0

x-1+y-1=x0+(-x0)=0

i.e. x+y=2

Because a-3a (2 3) + 5a (1 3) = 1 order a (1 3) = t

Get: t 3-3t 2+t = 1

T = 2 is obtained because b-3b (2 3) + 5b (1 3) = 5 let b (1 3) = f

Get f 3-3f 2 + 5f = 5

Find: f and then t+f= is the answer.

Answer: by a=, b=, and b really contains a

Introduced: The axis of symmetry x is less than 0 and the equation has a real root.

So: a is greater than 0 and a is less than or equal to -2 or a is greater than or equal to 2 to sum up: a>=2

is then a -4 0

There are -20 and a -4 0

The solution is a 2 and a >-2

Analysis: If b really contains a, then a is a true subset of b.

First of all, there is a true subset of the set, so first consider whether a is and then since the elements in b are all negative, the elements in a must be 2 negative numbers or 1 negative number.

So consider the relationship between the sum of the two roots and the product of the two roots, and discriminate 0 at the same time