1. Assuming such a triangular routing condition exists, abc=ab+bc+ca.

So c(ab-a-b)=ab.

C (A- 1) (b-1) = ab+c = (a-1) (b-1)+a+b+c-1,that is, (a-/kloc-).

A, b and c are positive integers, a

Then bc=2b+2c, then, (b-2)(c-2)=4. To solve this equation, only B is satisfied.

So a & gt=3, b & gt=4 and c & gt=5.

(a-1) (b-1) (c-1) > = 6 (c-1) > a+b+c+3 (c-2) > a+b+c-/kloc-

So such a triangle does not exist.

2. Let the number of sides of a polygon be n and the internal angle =(n-2)* 180 degrees/n.

The number of other edges is 2n, and the internal angle is (2n-2)* 180 degrees /2n.

(n-2)/n:(2n-2)/2n=3:4

8n- 16=6n-6

n=5

2n= 10

Their sides are 5 and 10 respectively.

3. Each vertex has only one regular polygon.

The sum of the three internal angles around the vertex is 360 degrees.

Assuming that these three angles are ∠A, ∠B and ∠C respectively, the sum of the external angles of the three angles

=( 180-∠A)+( 180-∠B)+( 180-∠C)= 540-360 = 180

External angle =360 degrees/number of sides

∴360/a+360/b+360/c= 180

∴ 1/a+ 1/b+ 1/c= 180/360= 1/2

4.ax+2y=2 x-y=-3b has an infinite group solution, then two straight lines coincide.

That is, a/1= 2/(-1) = 2/(-3b)

a=-2，3b= 1，

Equation 2ax+5=6b is -4x+5 = 2 and x = 3/4.

5.2x+ay= 16 1 formula

X-2y = 0 2 formula

1 formula -2 formula *2

2x+ay-2x+4y= 16

(4+a)y= 16

Y= 16/(4+a)>0, and y is an integer.

So 4+a= 1 or 2 or 4 or 8 or 16.

So a=-3, -2, 0, 4, 12.

6. Solution: subtract Chen Jia's score from the full score, 100-77=23.

So Chen Jia * * * deducted 23 points.

Doing the wrong question 1 is equivalent to deducting 10+6= 16, and doing the right part is equivalent to deducting 10-3=7.

So there is only one possibility: 23= 16+7.

So Chen Jia got 8 questions right, but did 1 wrong, and did part of 1 wrong.

7. Solution:

2x-a & lt； The solution set of 1 is x.

x-2b & gt； The solution set of 3 is x & gt3+2b.

Inequality group {2x-a

So (A+ 1)/2 = 1, and A = 1.

3+2b=- 1，b=-2

(a+ 1)(b- 1)

=( 1+ 1)(-2- 1)

=2*(-3)

=-6

I hope it helps you.