Let s get some of the 2nd difficult math problems as long as the questions don t have answers! ）

Question: The image of the primary function y=(2m-1)x+(m-5) does not pass through the second quadrant, and the value range of m is obtained.

Because the image of the primary function y=(2m-1)x+(m-5) does not pass through the second quadrant.

So 2m-1>0 and m-5=

1.On the right side of the equation is the calendar acre 1, and there are three possibilities, which need to be discussed by classification

The first possibility: the index is 0, and the base is 0;

The second possibility: the base is 1;

The third possibility: the base number is -1, and the exponent is an even number Answer: Solution: (1) When x+3=0, x2+x-1≠0, the solution is x=-3;

2) When x2+x-1=1, the solution is x=-2 or 1

3) When x2+x-1=-1 and x+3 are even, x=-1 is obtained

Therefore, all integer solutions of the original equation are -3, -2, 1, and -1, a total of 4

Therefore, choose b 2There are two cases in this problem, the minimum value of r is the radius of its circumscribed circle According to the sinusoidal theorem, 2r = ab sin

c = 15 12 13 = 65 4, r = 65 8, and if abc is an obtuse triangle, then the radius of the circle that can completely cover abc is half of the longest ab Answer: Solution: 2r= ab sinc= 15 12 13= 65 4,r= 65 8;

And if abc is an obtuse triangle, then the radius of the circle that can completely cover abc is half of the longest side ab, so r= 15 2=

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

1) x 2 + x - uncurved bi is equal to 0, x + 3 = 0 then x = -32) x 2 + x - 1 = 1

x^2+x-2=0

x=-2,x=1

3) x 2 + x - 1 = -1 and x + 3 is an even number.

x^2+x=0

x=0,x=-1

So x=-1

So x=-3, x=-2, x=-1, x=1

There are 4 of them. 2.Let ad be x, then db is 15-x according to the Pythagorean lead theorem.

13 2-x 2=12 2-(15-x) 2 solution x=25 3

13^2-x^2=896/9

cd = root number 896 9 approx.

In the same way, we can also think of ac = 13 and ad = 12 as being buried in diameter and the result is about and so that when ab is on the diameter, the shortest radius of the required circle is 10

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

1) x 2+x-1 is arguing with 0, and the fierce spike makes x+3=0 then x=-32) x 2+x-1=1

x^2+x-2=0

x=-2,x=1

3) x 2 + x - 1 = -1 and x + 3 is an even number.

x^2+x=0

x=0,x=-1

The branch is given x=-1

So x=-3, x=-2, x=-1, x=1

There are 4 of them.

There are x people. then one negative, x **.

So a total of yuan.

Less than $1 per person.

So the total price is less than yuan.

Therefore<> resistant to punch x > god world tour mu limb 3

So there are at least 6 people.

According to the title.

ec=dcac=bc

eca≌△dcb

1+ adc = 2+ adc = 180° i.e. e + adc = 180°

2) When point D is in the center of AB.

Quadrilateral AECD is

Because ACB is an isosceles right triangle.

Point D is at the center of when point D is in the center of AB, i.e., CD AB, E+ ADC = 180°

i.e. e=90°ec=cd

When point D is in the center of AB.

Quadrilateral AECD is

The first question seems to be missing a condition (isosceles).

Question 2: You add up the first and last ones to know how to do it.

Originally, junior high school algebra and geometry were my strengths, but I haven't seen them for more than ten years, and I've almost forgotten them.

(1).

28 is the mystery number: 8*8 - 6*6 = 282012 is the mystery number: 504*504 - 502*502 = 2012(2).

Proof : 2k+2) The value of 2 - 4k 2 is a multiple of 4 = 4k 2 + 8k + 4 - 4k 2 = 4 + 8k

4(1+2k)

4 (1+2k) is a multiple of 4.

The definition of the mysterious number is as stated in the title:

If a positive integer can be expressed as the difference of the squares of two consecutive even numbers, then the positive integer is mystical (2x) 2 - 2y) 2 (x > y) mystical numbers are even, and the square difference of two consecutive odd numbers (taken as a positive number).

k-1) 2 - k2 = 2k + 1 is an odd number.

The square difference of consecutive odd numbers (taken as a positive number) is not a mystery number.

Both are 2. Each item is a quarter of a maximum, except for one.

Question: The image of the primary function y=(2m-1)x+(m-5) does not pass through the second quadrant, and the value range of m is obtained.

Solution: Because the image of the primary function y=(2m-1)x+(m-5) does not pass through the second quadrant.

So 2m-1>0 and m-5=<0, i.e. 1 2