High school math questions for solving. High school math questions

According to the title, the coordinates of the intersection point: a(4,0); b(0，3)s△aob=1/2×4×3=6

To minimize cd, then cd oa, i.e., m oa, let m be: x=a, (04 rounded) 12-3a) 4=(3 2) 2

i.e. the minimum value of cd is (3, 2), 2

As shown in the figure, it is known that the straight line l:3x+4y 12=0 and the positive semi-axis of the x,y axis intersect at a and b points respectively, and the straight line l1 and the line segment ab,oa intersect at c and d respectively and bisect the area of AOB (1) Find the area of AOB; (2) Find the minimum value of cd - high school mathematics - Rubik's cube.

With answers and detailed explanations, I hope to help you.

The knowledge points examined by the 2x root number 3 are basic inequalities.

Let d(a,0), c(b,(12-3b) 4) finally obtain: 1 2(4-a)*1 4(12-3b)=1 2*1 2*3*4.........1）

Then by the basic inequality: 2ab<=a 2+b 2, both sides open the root number at the same time.

Think of the left side of (1) as the left side of the inequality.

Then further calculations can be done, and the result is 2 times the root number 3

Summary. If n is 2 and m is 1, then a cannot be negative.

Hello, can you describe the topic in detail? It would be nice to be able to take pictures, thank you.

I need you to send me a question so I can help you answer it.

If n is 2 and m is 1, then a cannot be negative.

Right. But why is the definition only given a 0, I thought there was only a 0 at the beginning, but why do we come out with so many a?

Did the title say the range of m and n?

The mn he gives is all positive integers.

So mn can be 1 and 2, and if the base is negative, the error is made.

And the title says positive.

Eh,What does this mean to imitate imitation?,That is, I just gave you the only big difference ** is right.,When I do the question, I can write it.,It should be a 0 in line with all the circumstances.,It's not comprehensive.。 And we should pay attention to the situation that the denominator in the previous index is 0, because this is also meaningless, and it doesn't work, there are a lot of problems, thank you.

The positive score given by the question does not need to consider the negative number.

The condition in the parentheses given later is to explain what a positive fraction of a positive number is.

What do you mean, then the one I sent you can't be used, you are worthy of the positive score of the positive number given by Xuxuan, why is it a positive score, isn't there still a hail of negative reputation, and what does Pei Feng mean by the question conditions.

The title says the meaning of prescribing the exponential power of positive fractions for positive numbers.

The condition given later is to explain the specific form of the positive fraction of a positive number.

As for negative scores, there are some, but this question is not covered at all.

a is the positive number in the question, and m n is the exponential power of the positive fraction.

As for the negative exponential powers you encounter in other questions, you need to distinguish the range of a on a case-by-case basis.

Your mind is solidified.

The fact that x does not exist in p does not mean that p does not exist or is not true, because p is a set, and if the condition for determining the elements of the set is not true, then the set is an empty set, where p is.

So the guy upstairs is right that p is an empty set, and an empty set is a true subset of any non-empty set, so p can deduce q

In other words, an empty set is the smallest set in scope, and without an element, it is natural to launch a larger set, i.e., an arbitrary non-empty set.

1. Since p is an empty set and q is not an empty set, this question can only say that the set p is a subset or a true subset of the set q, and the sufficient and necessary conditions cannot be discussed.

2. What you call a small range and a large range of launch does not refer to a set, but to an interval. There is a difference between a set and an interval, and the biggest difference is that an empty set cannot be represented by an interval. The reason for your problem is that the set p exists, but it cannot be represented by intervals, so there is no problem of scope and non-scope.

Since x does not exist, p is an empty set, and an empty set is a true subset of any non-empty set, and p is a true subset of q.

In words, where x belongs to the empty set, x belongs to {x, x<3}, so his inverse negative proposition is annoying, x does not belong to {x, x is greater than or equal to 3}, and x does not belong to the empty set. It seems paradoxical.

If the radius of the ball is the same as the radius of the cylinder, the maximum number of balls is the same as the integer multiple of the height of the cylinder and the diameter of the ball, and the diameter of the ball is 2*3=6

The height of the cylinder is 4+2 3=4+, rounded to 7

Up to 7 small balls can be placed.

Let the radius of the ball be r, then according to the title (r+3) =(r-3) =(1+23-r), draw a right-angled triangle).If r=1, then the center of the ball is on a circle with a radius of 2 and the diameter is 2, so 6 can be placed.

You seem to have a problem with the title, the radius of the ball is the same as the radius of the ground, and the height is, this can only be put into one.

1) Solution: Let m(x,y).

k1=y (x+2) k2=y (x-2) because: k1k2=m

So: k1k2=y (x+2) y (x-2) simplify: x 4-y 4m=1 So the curve equation is: x 4-y 4m=1

2) If m 0, then x 4-y 4m=1 is hyperbola.

If m 0 and m is not -1, then x 4-y 4m=1 is an ellipse.

If m=-1, then x 4-y, 4m=1 is a circle.

3) m=-3 4, then x 4 + y 3 = 1 (ellipse) straight line a1p: y = k1x + 2k1

with a2p: y=k2x-2k2

Let x= y2=2k2

k1k2=-3/4

p1p2=l6k1-k2l=9/2lk2+1/k2l.k2+1 k2 = 2 or k2+1 k2 =-2So 9 2lk2+1 k2l = 9 (when k2 = 1, or -1 is equal to the equal sign, i.e. y=x-2 or y=2-x).

The minimum length of P1P2 is 9

1, the proposition "If there is constant f(x) f(x) f(x 1) for a function y f(x) that defines the domain r as r, then f(x) is an increasing function, and the proposition is a true proposition because.

f(x) increases on r as the independent variable x increases.

It conforms to the characteristics of the increase function.

So f(x) is an increasing function.

2，y=2^(x+1)

It's a true function, for unclear reasons, but its domain is r, and it always has f(x)y=2 to the nth power.

Question 1: True propositions are increasing functions.

Question 2: y=2 to the power of x+1.

The first proposition is true because x+1 x and f(x+1) f(x).

y=2^(x+1)

Because t [1,2], there must be a value of y satisfying the equation (y-2) t-3y-1=0 so that t has a solution.

LZ got it? If you don't understand, I'll explain it to you.

Have you read the original question? The original question might be:

Find the range of y=(2t+1) (t-3) (1<=t<=2). ”

Maybe the original problem was also changed by substituting variables, so you didn't notice the range of t=f(x), which may be [1,2], which is 1<=t<=2

According to the solution process, the question less condition t belongs to (1,2), you can see it! In fact, killing chickens with a knife and exchanging T-3 for yuan is much better than the main yuan method!

There is something wrong with this method. First of all, the problem does not tell you the definition domain of t, so we get that t is not equal to 3, where does t have a solution in [1,2]?

x (1,+ then x (1,+.)

Obviously, when a=1, a x is satisfied.

Obviously, in order to make the inequality of any x on [0,m] constant, that is, such a should make the value of the sine and cosine function on both sides of a, and then draw the image of the sine and cosine function and find that the corresponding values of the sine and cosine on [0, 4] are only on both sides of a value, and it is not possible to exceed this range, so a is chosen

Is it to find the maximum value of m? A b options are also acceptable, but if you want to find the maximum value of m, you should choose c