Two high school math questions are slightly difficult and two high school math questions hard .

Easy, first of all, f(x)=f(-x)So f(k*2 x) + f(2 x-4 x-2)<0 is equivalent to f(k*2 x) f(2 x-4 x-2) f(4 x+2-2 x).

Since we know that f(x) is a monotonically increasing function defined on r, k*2 x 4 x+2-2 x is k<4 x+2-2 x 2 x, and to satisfy the proposal, as long as the right side of the inequality is the minimum value, the derivative can be used to find the minimum value

In the second way, let an be y, and the two-element one-dimensional equation is obtained, and the general term formula can be obtained by cross multiplication.

f(k*2 x)+f(2 x-4 x-2)<0. f(k*2^x)<-f(2^x-4^x-2)=f(4^x-2^x+2)

Because f is a monotonically increasing function.

k*2^x<4^x-2^x+2

Namely. 2 x) 2-(k+1)*2 x+2>0 because 2 x>0, so.

1) When k+1=<0, i.e. k=<-1, the above equation is constant.

2) When k+1>0, i.e., k>-1.

2 x) 2-(k+1)*2 x+2=[2 x-(k+1) 2] 2+2-[(k+1) 2] 2>0 to be constant.

Since [2 x-(k+1) 2] 2+2-[(k+1) 2] 2>=2-[(k+1) 2] 2

The equal sign holds if and only if 2 x=(k+1) 2

Therefore, it is necessary to (2 x) 2-(k+1)*2 x+2>0 constantly.

2-[(k+1) 2] 2>0 is required

The solution gives -1, so the range of value of k is .

k< (4 root number 2)-1

f(k*2 x)+f(2 x-4 x-2)<0f(k*2 x)<-f(2 x-4 x-2) odd functions. f(k*2^x)0

2 x-(k+2) 2] +2-(k+2) 4>0 as long as 2-(k+2) 4>0 is inequality.

i.e. (k+2) <8

2 22-2 2 Unclear title. a(n+1)=[3a(n-1)]/(-an+3)??

Knowing that three positive numbers a, b, and c satisfy a b+c 3a, then (b-2c) what is the minimum value of a?

Solution: Find the minimum value of (b-2c) a, first change the element and then find the defined domain, the linear programming can be set b a=x c a=y [xy is greater than 0] and then divide a b+c 3a by the above two formulas to get 1 x+y 3 a, b, c are three positive numbers, divide 3b a(a+c) <=5b by a to get 3x -1 y 5x -11 x+y 3

3x²-1≤y≤5x²-1

x,y o objective function z=x-2y

Linear programming finds the minimum.

1.Let t=xy. Mean inequality.

The maximum value is known by the form f(x,y) at the boundary of t=xy. The range of t=xy can be obtained by combining the equation (1-xy) =2(1-x)(1-y) with the mean inequality, when x=y=(root number 2)-1, xy takes the maximum value, and f takes the maximum value.

2.Let b a=x, c a=y, and the objective function is x-2y

From the constraints a b+c 3a,3b a(a+c) 5b, 3x 2-1<=y<=5x 2-1,1<=x+y<=3, and then the feasible domain is drawn by nonlinear programming, and the minimum value can be obtained at the boundary point. Do the rest yourself.

1.True, both. By definition, a rhombus is a flat quadrilateral with equal sides.

A square is a flat quadrilateral with equal sides and right angles. Of course, there is a difference between the two, if a figure is a square, then it must be a diamond. But if a figure is a diamond, it doesn't have to be a square.

2.The second question is that what you said is right, it is not a sufficient condition, but a necessary and insufficient condition.

1) Statement 2 is correct, only one diamond with an inner angle of 90 degrees is a square, the others are not. 2) Your teacher is wrong, this is not a sufficient condition, there are many examples, such as 1, 2, 16, 32, it is not a proportional series, but it is in line with AD=BC

1. Yes, all squares belong to diamonds, but all diamonds do not necessarily belong to squares, which is understood through the definitions of both.

2. Your understanding is correct, you can give a counterexample to show that this proposition is a false proposition, and the existence of 0 that your teacher said can also ensure that this ad=bc is true, but 0 cannot be used as a divisor. This is not true.

This problem is solved using a special straight-line method. (After this sentence comes out, it is not a problem) by the question. It is discussed in two cases.

1）a<0

In this case, take the x-axis as a special straight line, i.e., the straight line y=0 (over the point m) as an example. At this point, p and q coincide. 1 MP squared + 1 mq squared = 2 A squared.

However, for any straight line y=k(x-a) (k is not equal to 0), mp and mq are larger than those at y=0. Therefore, 1 mp square + 1 mq square is not a fixed value. Abandon.

2）a>0

In the same way, make 2 special straight lines that cross the point m. Let's take the lines y=0 and x=a.

When y=0, when 1 mp squared + 1 mq squared = 2 a square x=a, the parabola is substituted to obtain y=+- root number 2pa; In this case, mp=mq=root number: 2pa, 1 mp squared + 1 mq, squared = 1 pa

In both cases, the value of 1 mp squared + 1 mq squared is the same. Therefore a=2p

There are three cases where the position of 2 is even, where 2 is equivalent in the first and third:

2 in the first place: 2 4 6 or 2 6 4 2 equivalents.

Put 1 first: If 1 is on the left side of 2, there are two ways to put it.

If 1 is on the right side of 2, choose one to put between, and the other can be placed on both sides, four cases.

Yes: 2 * 2 + 4) = 12;

2 In the third place, the same is true, there are 12 kinds;

2 in the middle: 4 2 6 or 6 2 4 2 equivalent.

1 has two places to put, both places are equivalent, 1 can put one of them (2 kinds), the remaining one needs to put 3 or 5 (2 kinds of cases), and the remaining one can be put at both ends (2 kinds) after putting it

Therefore, 2 is in the middle: 2 * 2 * 2 * 2 = 16 In summary, there are 40 kinds of total, and you can reply to the second question if you need a more detailed process: sina + sinb = root number 2) sinc It can be seen that a + b = root number 2) c

There is a + b + c = root number 2

CLP: c = 2 - root number 2) is the ab side length.

1.Solution: f(9)=f(3*3)=f(3)+f(3)=2f(a)>f(a-1)+2=f(a-1)+f(9).

a>0, a-1>0 gets a>1

f(a)>f(9a-9), and f(x) is an additive function defined on (0,+).

a>9a-9, i.e. a<9 8

So, 1=0 (because of 0).

a>=3 2 or a"Oak cherry = -1;

Liang Zhencong (x-2a) 2=4a 2-2a-6 gives x=2a + root number (4a 2-2a-6) or x=2a - root number (4a 2-2a-6).

2a+ (4a, 2-2a-6), <0 or 2a-root (4a, 2-2a-6), <0

The solution is -3, so the value range of the actual number a is: a -1

From the condition -1 2x+1 2)(x-1 3)<0, the parentheses of the inequality are opened and simplified so that the inequality sign is the same as the inequality sign of the original inequality, and the constant term is the same. You can get the values of a and b. From this, we can calculate the solution set of x 2+bx+9<0 in the future.

I hope you can solve it by yourself.

The solution set for ax 2+bx+1>=0 is -1 2b a=-1 6,b=-1

The solution set of x 2-x+9<0 is an empty set.

Knowing the solution set, you can use Vedder's theorem, compare the magnitude of 2 values, and then sit with the sign to find the coefficient.

Then use Veda's theorem to compare the magnitude of 2 values to obtain the solution set.