An 8th grade math problem

Spreads... Cross oh ...

Kilometer? Highway...

Sweat... There are two situations when three highways intersect, one is to cross one point, and the other is to cross three points, that is, to form a triangle. There is no picture, so I'll talk about it separately.

If it is at one point, it is not solved, unless the cargo transfer station is built to the intersection point, then it is in the middle of the road, which is impossible.

If you are enclosing a triangle, there are four addresses to choose from: the inner (intersection of the inner bisector) and the outer center (the intersection of the outer bisector).

The distance from the cargo transfer station to the three highways is equal, and the three highways are A, B, and C. They are intersected at three points: a, b, and c. Let A and B be handed over to C. The cargo transfer station is assumed to be built at point p.

Then since the distance from p to the straight line a and b is equal, the distance from one point outside the two intersecting lines to the two straight lines is equal, and it can only be a point on the bisector of the intersection angle. p is located on the bisector of the straight line a, b. A, B intersect C, there are four angles, and point P is located on the bisector of these four corners.

Of these four angles, two groups are two rays that are opposite to the vertex and are bisector lines of the two angles of the vertex on the same straight line and in opposite directions to the same vertex. So, point p is on two intersecting lines (let it be m,n), which are actually perpendicular to each other.

In the same way, p is located on the bisector of the intersection of the straight lines b and c, and is also located on two straight lines.

p is located on the bisector of the intersection of the lines a and c, and is also located on two straight lines.

So there are 6 straight lines, and if you plot it, you will find that they intersect at 4 points. One of the points is located inside the triangle ABC, this point is the intersection of the angle bisector of the angle ABC, the angle ACB, and the angle BAC, and these three lines intersect at the same point, which is mathematically called the heart, which is the center of the inscribed circle. Since this point is located on the bisector of the three angular angles, it is clear that the distance to the three straight lines is equal.

There are also 3 points, which are three straight lines of the bisector of the outer corner of the triangle, and three points formed by the intersection of two pairs. We call this the outer mind.

Take the angle abc and the angle acb, the outer angle bisector of these two angles intersects at a point, then since this point is located on the bisector of the intersection angle of the straight line ab and bc (i.e., the complementary angle of the angle abc), the distance to ab and bc is equal. In the same way, the distance from this point to AC and BC is equal, that is to say, the distance from this point to the three straight lines is equal, which is in line with the topic.

The other two points are the same, as you can see from the following figure.

2.(m²-2m+1)+(n²-4n+4)=0m=1,n=2 mn=2

3.From the question (1 x x) 2=1 x +x -2=41 celery belt x +x =6

Original = 04=2a(x²+x+1)

5.(a+b+c) 2=a +b +c +2(bc+ac+ab)=0, i.e., 1 +2(bc+ac+ab)= 0bc+ac+ab =-

A+B+C) 2 -4 (BC+AC+AB) = 3

2x to the power of 2 - 2ax + 3x - 3a = 0

2x*(x-a)+3(x-a)=0

2x+3)(x-a)=0

So 2x+3=0, so x=-2 3

Yes, of course! Just substitute x = minus 2/3 directly, if the equation is true, if it is not true, it is not.

(a-4) x 4-a can become x 4-a (a-4) = -1, so that the direction of the unequal sign is not changed.

A-4 must be greater than zero.

And 4-a is not equal to zero.

So a is greater than 4

(a-4)x＞4-a

a-4)x＞-(a-4)

Divide by a-4 on both sides

The solution set is x>-1

The direction of the unequal sign does not change.

So the division of a-4 is a positive number.

then a-4>0

a>4

（a-4）x>-(a-4)

a-4)x+(a-4)>0

a-4)(x+1)>0

x>-1, so (x+1)>0

If the equation holds, (a-4) >0

Hence a>4

If a-4>0 then x>(4-a) (a-4)=-1, so a>4 satisfies.

If a-4<0 then x<(4-a) (a-4)=-1 does not satisfy the sum a<4 (the emphasis is on the question of whether the direction of the unequal sign changes).

Using the Pythagorean theorem to show that the left and the right are 90 degrees, that's it.