If the equation m minus 1 divided by x minus 1 minus x plus 2 divided by x minus 1 equals 0, then M equals 0

If the equation m minus 1 divided by x minus 1 minus x plus 2 divided by x minus 1 equals 0, then M equals 0

four

If M is the solution of the equation | 2000-x | = 2000 + | x |, then | m-2001 | is equal to ()
A m-2001 B -m-2001 C m+2001 D -m+2001
2. If the equation | 2x-3 | + M = 0 has no solution, there is only one solution for | 3x-4 | + n = 0, and there are two solutions for | 4x-5 | + k = 0, then the size relation of M, N and K is ()
A、M>N>K B.N>K>M C.K>M>N D.M>K>N
3. There are () values of integer x suitable for the relation | 3x-4 | + | 3x + 2 | = 6
A. Natural numbers with 0 B.1 C.2 D. greater than 2
4. If the equation | x + 1 | + | X-1 | = a about X has real roots, then the value range of real number a is ()
A.a≥0 B.a>0 C.a≥1 D.a≥2

The solution of the first problem is: square the left and right sides at the same time
- 4000x = 4000 | x |, it is obvious that x is less than or equal to 0, that is, M is less than or equal to 0
The answer is d
The solution of the second problem is: move m, N and K to the right of the equation. From the property of absolute value, we can know that: | x | = y has no solution Y0, so the answer is: a
The solution of the third problem is: because the absolute value is greater than or equal to 0, you only need to verify the seven integers 0-6, for example: | 3x-4 | = 0, | 3x + 2 | = 6 to see if there is a solution
The answer is: C (x = 0,1)
The solution of the fourth problem is: the geometric meaning of | x + 1 | + | X-1 | represents the sum of the distances from a point on the coordinate axis to - 1 and 1, so the minimum value is: 2, that is, a ≥ 2
The answer is: D

If M is the solution of the equation | 2000-x | = 2000 + | x |, then | m-2001 | is equal to ()
A. m-2001B. -m-2001C. m+2001D. -m+2001

According to m is the solution of the equation | 2000-x | = 2000 + | x |, so | 2000-m | = 2000 + | m |, when m ≥ 2000, M-2000 = 2000 + m, contradiction; when m ≤ 0, 2000-m = 2000-m, constant; when 0 < m < 2000, 2000-m = 2000 + m, solution: M = 0 does not conform to the meaning; so m ≤ 0, | m-2001 | = - M + 2001