If the fractional equation x-3 3-2x + 3-x 2-mx = - 1 has no solution, find the value of M Time is short

If the fractional equation x-3 3-2x + 3-x 2-mx = - 1 has no solution, find the value of M Time is short

Through differentiation
x(m-1)=2
When m = 1
The equation has no solution

When m=______ The fractional equation 2x + m with respect to X X − 3 = - 1 has no solution

If we remove the denominator of the equation, 2x + M = - x + 3
X = 3 − M
Three
When the denominator x-3 = 0, that is, x = 3, the equation has no solution
So 3 − M
When 3 = 3, the equation has no solution
The solution is: M = - 6

If the equation 2x-3 + M = 0 on X has no solution There is only one solution for 3x-4 + n = 0 and two solutions for 4x-5 + k = 0 A.m=k

Because there is no solution for | 2x-3 | + M = 0, and x-3 | ≥ 0
Therefore, when m > 0, the original equation has no solution
Because there is only one solution for | 3x-4 | + n = 0, and | 3x-4 | ≥ 0
So when n = 0, the original equation has only one solution
Because | 4x-5 | + k = 0 has two solutions, and | 4x-5 | ≥ 0
Therefore, if K < 0, the original equation has two solutions because | 4x-5 | can be taken as ± K
Because m > 0, n = 0, K < 0
So the relationship of M, N, K is m > n > K

If the equation (x + 3) / (2x + 1) = 4 / (m-2) about X has no solution, then M =?

When m is not equal to 2, multiply by open to obtain (m-2) * x + 3m-6 = 8x + 4, then M = 10
When m = 2, there is no solution

When m =, the equation X-2 of X + 1 = - M has no solution

m=-2

On the fractional equation of X (x-3 / 3-2x) + (3-x / 2 + MX) = - 1

(3-2x)/(x-3)+(2+mx)/(3-x)=-1
(3-2x)/(x-3)-(2+mx)/(x-3)=-1
(3-2x)-(2+mx)=-(x-3)
3-2x-2-mx=-x+3
-x-mx=2
-(1+m)x=2···········①
To make the original equation have no solution, let x = 3 (x = 3 is the augmented root of the equation)
-(1+m)*3=2
m=-5/3.
In addition, for formula ①, if 1 + M = 0, then x = 0, but the two sides of the original equation are not equal, so x ≠ 0. Therefore, to make the original equation have no solution, only make 1 + M = 0, that is, M = - 1,
To sum up, when m = - 1 and - 5 / 3, the original equation has no solution

The fractional equation of X (3-2x) / (x-3) + (2 + MX) / (3-x) = - 1 has no solution, and the value of M is calculated

The original equation is as follows:
(3-2x)/(x-3) - (2+mx)/(x-3) + (x-3)/(x-3) = 0
(3-2x-2-mx+x-3)/(x-3) = 0
[(-m-1)x-2]/(x-3) = 0
Discussion:
(1)
When the molecule of the formula is a constant which is not equal to zero, the original equation has no solution
-m-1=0
m=-1
(2)
When the numerator is the multiple of nonzero constant term of denominator, the original equation has no solution. Let this constant term be C (C ≠ 0), that is:
[(-m-1)x-2]/(x-3) = C
Then:
(-m-1)[x-2/(-m-1)] = C(x-3)
2/(-m-1) = 3
m=-5/3
To sum up: when m = - 1 or - 5 / 3, the original equation has no solution

If the fractional equation x + 2x of 1 = x + 1 / M has no solution, what is the value of M?

2x/(x+1)=m/(x+1)
2x=m
When m = - 2, x = - 1
Because x = - 1 denominator x + 1 = 0 is meaningless
So when m = - 2, the fractional equation x + 1 / 2x = x + 1 / M has no solution

It is known that the solution of X equation 3 (M + x) = (1 + 2x) / 2 is 2 larger than that of X equation x (M + 1) = m (1 + x)

Solve the first equation: 6x + 6m = 1 + 2x
X = (1-6m) / 4
The second equation: x = M
Then: (1-6m) / 4-m = 2
The solution is: M = - 7 / 4
The principle is OK, if the result is not right, calculate it by yourself

It is known that the solution of equation 3 (x + 3) - 1 = 2x is the same as that of equation 3x + M = 4 / m-27 with respect to X

3(x+3)-1=2x
3x+9-1=2x
x=-8
Substituting into the second equation, we get the following results:
3*(-8)+m=(m-27)/4
-24+m=(m-27)/4
-96+4m=m-27
3m=96-27
3m=69
m=23