Rush to death! A math problem for the second year of junior high school!!

2x—m is less than or equal to 0

2x≤m x≤m/2

3≤m/2<4

6 m<8 or 2x—m 0

Get x m 2

And because there are only three positive integer solutions, then the smallest three positive integers 1, 2, and 3 can be taken because 3, so m 2 should be greater than or equal to 3 (if m 2 is less than 3, then x will not be able to take 3).

m 2 should also be less than 4 (if m 2 is greater than or equal to 4, then x can be taken to 4, which is contradictory to only 3 positive integers).

So 3 m 2<4

So 6 m 8

Because 2x-m 0, so 2x m, so x m 2, and because the positive integer solution of the inequality 2x-m 0 is 1,2,3, so 3 m 2 4, so 6 m<8

Finishing x is less than or equal to half m

So 8 is the boundary of m.

This is not true when m is 8.

m is less than 8

2x-m≤0

2x≤mx≤m/2

The positive integer solution of 2x-m 0 is.

Get 3 m 2 4

The value range of m is 6 m 8

2x-m≤0

2x≤mx≤m/2

The positive integer solution is.

3≤m/2≤4

The value range of m is 6 m 8

Method 1:

Method 2: (Concise).

As shown in the figure below:

Children, you have to do it yourself.

Because eb=ec. So the angle EBC = angle ECB, because the angle ABE is equal to the angle ACE, so the angle ABC= angle ACB, so the edge AB=AC, and because EB is equal to EC, the angle ABE is equal to the angle ACE, so the triangle ABE is congruent with the triangle ACE, so the angle BAE is equal to the angle CAE

This question seems to be based on basic training.

Figure is not clear, eb=ec

EBC = ECB (equilateral equilateral equiangular angle).

and d is the midpoint of BC.

abe= ace (known).

ABC = ACB (Equal plus Equal).

AD bisects BAC (isosceles triangle three lines in one).

The carrying principle, the distance from the point on the corner bisector sock to both sides of the corner is equal, and it is applied twice.

2. r= cde+ c, i.e. cde= r- c= r- b= ade- b= adc- cde- b, transform it into 2 cde= adc- b= b+ a- b= a, so cde is related to a, so b.

I think you should have copied the wrong question.

The title should be:

When m= is the difference between the fractional equation for x (2x+m) m-3=-1 without a parenthesis without a solution

If so, then:

2x+m=3-x

3x=3-m

x=1-m/3

We know. When x=3 x-3=0 there is no solution to this equation then 1-m 3=3

m/3=-2

m=-6If you think your question is correct, then:

2x(x-3)+m=3-x

2x^2-6x+m=3-x

2x^2-5x =3-m

x^ =3/2-m/2

x^( = -1/16-m/2

√（1/16-m/2）

x = (1 16-m2) + as above. x=3

√（1/16-m/2)+

（-1/16-m/2) =±（-1/16-m/2)=

Hypothesis (1 16-m2)=

m 2=m = hypothesis 1 16+m 2=

m/2 =3

m = 6 and when m = 6 (-1 16-m 2) is meaningless.

m=6 rounded. ∴m=

Answer: Because: (2x+m) (x-3)=-1, so:

2x+m=-x+3 So: 3x=3-m, the solution: x=(3-m) 3 is known:

x=3 is the root of the equation, so: (3-m) 3=3 so: m= - 6

The denominator is x-3

The numerator is 2x+m

Solve the equation to obtain: x=(3-m) 3

When x-3=0, i.e., x=3, the equation has no solution.

Then (3-m) 3=3, and m=-6

When m = -6, the fractional equation 2x+m x-3=-1 about x has no solution.

First, it is proved that the triangle AEC is all equal to the triangle ADB, and AE=AD can be obtained, and then the triangle AEF is fully equal to the triangle ADF. Then: eaf= daf

Because ABC is an isosceles triangle, CDF and BEF are congruent, so DF=EF and the angle ADB and the angle AEC are right angles, so AF is the angle bisector of angle A, i.e., DAF=EAF