High school math masters solve, sit back and wait for the answer, speed

Bringing (1,0) into gives (1-a)k+1+b-2a=0 for any kheng, so 1-a=0,1+b-2a=0 gives a=1,b=1

In this case, y=(k 2+k+1)x 2-2(1+k) 2 x+(k 2+3k+1)=(x-1)[(k 2+k+1)x-(k 2+3k+1)].

So the other root is x=(k 2+3k+1) (k 2+k+1)=1+2k (k 2+k+1).

So |ab|=Absolute value of the difference between the two sticks=|2k/(k^2+k+1)|

When k=0, |ab|=0

When k=0, |ab|=|2/(k+1/k+1)|=2/|k+1/k+1|, k+1 k2 or -2 from the checkmark function, so k+1 k+1 3 or -1, so |k+1/k+1|3 or 1, so |k+1/k+1|1, so |ab|=2/|k+1/k+1|≤2

In summary, |ab|The maximum value is 2

In the function y=(k 2+k+1) x 2-2(a+k) 2 x+(k 2+3ak+b), let y=0

k^2+k+1)x^2-2(a^2+2ak+k^2)x+(k^2+3ak+b)=0

x^2-2x+1)k^2+(x^2-a)k+(x^2-2a^2+b)=0

Since the above equation is true for any k, the coefficients in the polynomial in the above equation must be zero:

x^2-2x+1=0

x^2-a =0

x^2-2a^2+b=0

x-1)^2=0

x^2=ax^2-2a^2+b=0

x=1a=1

b = 1 function f(x) = (k 2 + k + 1) x 2-2 (1 + k) 2 x + (k 2 + 3k + 1).

Because one is 1 and the other is x2

by x2=(k 2+3k+1) (k 2+k+1)=1+2k (k 2+k+1).

x2-1|=|k/(k^2+k+1|=1/(|k|+1/|k|+1)≤1/3

ab|The maximum value is 1 3

a(1,0) is substituted into the function y=(k 2+k+1) x 2-2(a+k) 2 x+(k 2+3ak+b) to get (1-a)*k+(1+b-2*a 2)=0 for any k constant, so 1-a=0 and 1+b-2*a 2=0 gives a=b=1

The function y=(k 2+k+1) x 2-2(1+k) 2 x+(k 2+3k+1)=(x-1)*[k 2+k+1)x-(k 2+3k+1)]

ab|=|)*k^2+3k+1)/(k^2+k+1)-1|=|2k/(k^2+k+1)|=2/|k+(1/k)+1|<=2

So when k = -1, |ab|The maximum value is equal to 2

(2)｜b｜=4

b+a and a-b are squared respectively and then subtracted to each other to get 4 a b =32 and finally b =4

The rest of the questions don't seem to be complete.

6/(3-x)∈n

3-x is a divisor of 6.

3-x=-6，-3，-2，-1，1，2，3，6.

There are 8 values of x, so |p|=8

4,n is a set of non-negative integers, so the number that makes 3-x 6/6 belong to n is only 1, 2, 3, 6, that is, x is 0, 1, 2, -3, please believe the correct answer. Don't just look at the details.

(1) 4 52 = 1 13 (2) 13 52 = 1 4 (3) There are 6 prime numbers in each suit, which are 2, 3, 5, 7, 11, 13, so there are 24 prime numbers, and the probability is 24 52 = 6 13

Question 2, d

Question 3, k = -2

Question 4, =-3 2

This problem is a combination of numbers and shapes. I used the r instead of the Greek letter.

Combination of numbers and shapes.

Solution: The set a represents a circle with (0,0) as the center and 1 as the radius;

Set b represents y=丨x丨+ a square of four straight lines (this place is the key).

When it changes, it means that the square is changing.

When the four vertices of the square are the four intersection points of the circle and the coordinate axis, the minimum value of 1 is taken, and when the line is tangent to the circle, the maximum value is reached, and the root number 2 is obtained, so the range of values is [1, root number 2].

A greater than or equal to one or less than or equal to the root number two must be a positive number, because b is not an empty set.

If you have any questions, please feel free to ask, if there is no problem, please adopt it in time, thank you.

Use mean inequalities.

Because |x|AND |y|are all positive.

then [(x 2+y 2) 2] (x|+|y|2 i.e. (1 2) 2

So 2

And because =|x|+|y|

and x 2 + y 2 = 1

So 1 2

Greater than zero and less than 1, or greater than root number 2