I ask for help sorting it out in the first year of junior high school!Teach me math in the first yea

In fact, this is the problem of several important points in the triangle.

Clarify the concepts of these points to know their nature, and then solve the problem according to their nature.

Heart: That is, the center of the inscribed circle of the triangle, which is the point with equal distances to the three sides, because the points with equal distances to both sides of an angle are on the intersection of the angle bisector, so it is also the intersection point of the angle bisector. Therefore, the example you gave above is not valid, because the distance to the two sides of the corner is the right side of the right triangle, and the distance to the vertex is the hypotenuse of the right triangle, which cannot be equal.

Outer center: The center of the circumscribed circle of the triangle, the distance to the three vertices is equal, because the distance to the two ends of the line segment is equal to the point ** segment perpendicular bisector, so the outer center is the intersection of the triangle perpendicular bisector.

Center: The intersection of the midline of the triangle, which is the side opposite the vertex that is bisected by the line that connects this vertex.

As long as you have a clear understanding of the defining nature of these three points, the kind of questions you mentioned are basically the same. Do a few more questions and become proficient, and you should be fine.

The distance to both sides of the corner is the angle bisector, and the distance to the point is equal to the perpendicular bisector.

Normally, when you see the words that the three sides of the triangle are equal, you make the bisector of the two angles, and the intersection is the one you want.

If you see three points at equal distances, you're going to have to do a bisector.

Points on the angular bisector are equally spaced apart from each side. The bisector of the three inner angles of the triangle intersects with a point that is at an equal distance from the three sides.

(1) If a and d are set

The distance between the two places is:

x km, please list a about.

x. x=5*(50+70)/(5-3)

2) If the speed of the car is y kilometers per hour, make an equation for y.

y=(50+70)/(5-3)

x^2-y^2=8

x+y)(x-y)=8

x+y=4, so x-y=2

Simultaneous two equations.

Add to x=3

y=12009*2010*2011-2010^3=2010(2009*2011-2010^2)=2010[(2010-1)(2010+1)-2010^2]=2010[2010^2-1-2010^2]=2010*(-1)

x+1)(x^2+1)(x^4+1)(x^8+1)=(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)/(x-1)

x^2-1)(x^2+1)(x^4+1)(x^8+1)/(x-1)

x^4-1)(x^4+1)(x^8+1)/(x-1)=(x^8-1)(x^8+1)/(x-1)=(x^16-1)/(x-1)

1, x 2-y 2=(x+y)(x-y)=8 and x+y=4

So x-y=2

So x=3, y=1

3.The third question is what you are asking for?

Can be pushed out (x+y)*(x-y)=8, so x-y=2, so x=3, y=1

3。What do you want to ask?

Say something wrong!

6.If we know that the two square roots of a positive number are 3x-5 and x-7 respectively, then this positive number is .

3x-5) 2=(x-7) 2,9x 2-30x+25=x 2-14x+49,x 2-2x-3=0,(x+1)(x-3)=0,x=-1 or x=3,,when x=-1,3x-5=x-7=-8 (undesirable, rounded), when x=3 3x-5=4,x-7=-4, so this positive number is 16

7.It is known that OA OC and AOB: AOC=2:3, then the degree of BOC is .

As you can see, AOB = 60 degrees, AOC = 90 degrees, BOC = 30 degrees.

8.If 1=80°, the two sides of 2 are parallel to the two sides of 1, then 2= 80 degrees or 100 degrees. (Make a diagram below).

9.If the point A (1+m,2m+1) is known to be on the x-axis, then the coordinate of point A is (,0).

a is on the x-axis, the ordinate is 0, so m=, m+1=

10.Given the ab x-axis, the coordinates of point A are (3,2), and the coordinates of ab 5, then the coordinates of b are .

b(-2,2) or b(8,2).

11.If the distance from the point p(a-2,2a+3) to the two coordinate axes is equal, then a=

a-2|=|2a+3|, (a-2) 2=(2a+3) 2, a=-5 or a=-1 3

That's it, think about it for yourself. If you want to learn math well, you can't do it without doing a certain amount of problems.

12.Shift the point A(1, 3) to the right by 2 units, and then translate it downwards by 2 units to get the point B(A, B), then AB

13.It is known that the coordinates of the point p in the plane Cartesian coordinate system are (-1,3), if the plane Cartesian coordinate system is translated to the left by 3 units, and then by 2 units downward, then the coordinates of the point p after translation are

14.In the planar Cartesian coordinate system, where a(2,-2) is known, and a point p is determined on the y-axis, so that aop is an isosceles triangle, then there are a total of points p that meet the criteria.

15.Point p(a+5,a) cannot be in quadrant.

16 If there is a point p(x,y) in the plane Cartesian coordinate system, and if x = 0y is satisfied, then the point p is in.

18.Knowing that x=1,y=8 is the solution of the equation mx+y 1=0, then the square root of m is .

19.About the inequality of x (a+1)The set of solutions for x a+1 is x 1, then the range of values for a is .

20.If there are 3 solutions to a positive integer in equation 2x m 0, then the range of m is .

6.Because two numbers are square roots of each other.

So (3x-5) + (x-7) = 0, so x 3

15Second quadrant.

On-axis or -3 is not equal to -1

x<4

6. 3

7.60 or 120

9.(1/2,0).

10.(-2,2) or (8,2).

11.There are four answers, do the math yourself.

15.Second quadrant.

Shaft on 18+-3

You are too lazy to do it yourself, and you use labor for nothing.

Because the coefficient is 9, the number of times is 4

So m=9, n=4

Then mx to the power of 4 - ny to the power of m-n = 9x 4-4y 5 is the 5th degree binomial.

Solution: Set up the original X station.

3/5x+20):x=4:5

Solution: x = 100 units.

A: The mall originally had 100 televisions. There are now 80 televisions in the mall.

It turned out that there was a TV set x set.

Listing: [x-(2 5)x+20] x=4 5 solution x=100

So it turned out that there were 100 TV sets.

The mall originally had a TV set.

1-2/5)x+20:x=4:5

4x=3x+100

x=100 now has: 3 5 100 + 20 = 80 units.

According to the title, the ratio of 20 units to the original total is (1-2 5)-2 5=1 5

So there were 20 (1 5) = 100 TVs in the original and 100 (4 5) = 80 TVs in the existing TV.

There were originally X sets of televisions.

4:5 = (1- 2/5)x+20:x

4x = 3/5 x*5+20*5

4x = 3x+100

x = 100

1-2/5)x+20=80

There were 100 televisions.

There are now 80 televisions.

Set up the original X station:

x*（1-2/5)+20]/x=4/5

Solve x=100

There were 100 units originally.

Existing: 100*(1-2 5)+20=80 units.

Set the original x station, (3 5) x + 20: x = 4:5

x = 100 There used to be 100 units, but now there are 80 units.

Let the original x rule (3 5x+20): x=4:5

x = 100 units.

There are 80 units.

2, 3, 4, n - 3 quadrilaterals (n does not equal 3).

-a+1≠0，|a|+1=2, so that a=-1, and substituting to find the following equation is -7

Substituting x=4 into the equation gives a=-8 23

The following long string of pairs is combined to give 50*-a=-50a, which is -8 23 -50=400 23

Because of the quadratic trinomial, so |a|+1=2 and -a+1 is not equal to 0, so a=1 is rounded, so a=-1, bring the value of a into the following equation and wait for the answer.

Bring x=4 into the equation of x and solve it to get the value of a.

The following equation is equal to 50 (-a), a can be brought in to get the answer, please complete the calculation process by yourself, mathematical calculations are very important.

1) Since the original formula is a quadratic trinomial, (-a+1)≠0,|a|+1=2 to get a=-1

Finding the formula =-1 2-[3 2-3(-1-1 3)]-4+3=12) brings x=4 into the original equation 3a+4=a 8+3 to get a=-8 23;

Equation = -50a = 50 * 8 23 = 400 23

The answer is the same, which of the above is correct.

I can't help, and I ...... a quasi-junior life

Where was your first year of junior high school? I haven't been to school in a few years. The questions have been upgraded.