High school math problems, I hope someone can help explain them

First, "closed" is for operations and sets.

For example, the operation "addition" and the set of real numbers r, we all know that for any two elements in r, that is, two real numbers, the result of their addition operation must still be real numbers. At this point, we say that the set of real numbers r is closed to the "addition" of operations. Closure is actually the fact that the elements in the R set are "added" and the result is still the elements of the set, that is, these elements are enclosed in the set.

Now let's talk about how to verify that a set is closed to the operation @ in the problem.

1) Unocclusive: Only two elements need to be found, and the result of these two elements after the @ operation is not in the set.

2) Closure: Verification is still in the set for any (the word arbitrary is important) two elements that are still in the set after the @ operation.

You said that x1 and x2 are arbitrary, so mark 1 and 2 to distinguish the answer to the question, please see the figure below.

There are only four possible additions, subtractions, multiplications, and divisions. Closure is a kind of term, and you can also understand it as, @爱上a. @ in this question = multiplication, if"Multiply"If you fall in love with a, then x1 times x2 = x3, where x1, x2, x3 are all a.

And x1≠x2≠x3, so we can only assume.

Let x1 = m1 + 2n1, x2 = m2 + 2n2, x3 = m3 + 2n3, (is the subscript) then.

m1 + 2n1) (m2 + 2n2) = m3 + 2n3, that is, [m1m2 + 2n1n2] + 2 (n1m2 + n2m1) = m3 + 2n3, compared to both sides of the equation, it is found that as long as m1m2 + 2n1n2 = m3, n1m2 + n2m1 = n3The equation can be established.

And m1m2m3n1n2n3 are all integers, so m1m2+2n1n2=m3 and n1m2+n2m1=n3 are possible.

So take the fall in love with A.

For example, take m1=1, n1=1, m2=2, n2=2Substituting m3=6,n3=4At this time, x1 = 1 + 2, x2 = 2 + 2 2, x3 = 6 + 4 2, x1 times x2 = x3 holds, so multiply to fall in love with a, that is, @ to fall in love with a

The domain of the function f(x) is [-1,4], i.e. -1 is less than or equal to x less than or equal to 4

In the function f(2x+1), you should think of 2x+1 as x, that is, -1 is less than or equal to 2x+1 less than or equal to 4, and the x obtained from this is the domain of pin x in the function f(2x+1).

Do you understand? I don't know if I want to add my friend and explain it to you slowly.

The domain of f(closed tank x) is [-1,4], i.e. the range of the independent variable f(x) is [-1,4], and for f(2x+1), 2x+1 is the independent variable of f(x), so -1 is less than or equal to 2x+1 less than or equal to 4

The solution is as follows:

What about my answer?

Solution: Take the midpoint E of the AC edge and connect DE, D is the midpoint of BC, then DE is the median line DE= AB= ·4=2

ae=½ac=½·7=7/2

ac=7，ad=7/2

From the cosine theorem:

cos∠cad=(ae²+ad²-de²)/(2·ae·ad)cos∠cad=(ac²+ad²-cd²)/(2·ac·ad)(ae²+ad²-de²)/(2·ae·ad)=(ac²+ad²-cd²)/(2·ac·ad)

I tidy it up and get it. cd²=ac²+ad² -ac·(ae²+ad²-de²)/ae=7²+(7/2)² 7·[(7/2)²+7/2)²-2²]/(7/2)

cd=9/2

bc=2cd=2·(9/2)=9

The length of the BC is 9.

The answer doesn't seem quite right, but the process should be fine, so let's see for yourself.

Only option 3 is correct, which one doesn't understand?

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

Correct ......

1 Patently wrong. 4 should be an extreme value of |a^2-b|

The second one is not right, it can only be judged that b=2, and the third is also wrong, because a2>=0, so b>=0, the axis of symmetry is a, but it cannot be judged to be positive or negative, and in a-positive infinity, it decreases first and then increases, the third is wrong, sorry, I think it's all wrong.