If the equation (AX + 1) / (x-1) - 1 = 0 about X has no solution, then a is I know that a = - 1, but if we divide the equation into x (A-1) = - 2, then a = 1 is no solution?

If the equation (AX + 1) / (x-1) - 1 = 0 about X has no solution, then a is I know that a = - 1, but if we divide the equation into x (A-1) = - 2, then a = 1 is no solution?

(ax+1)/(x-1 )-1=0
(ax+1-x+1)/(x-1)=0
[(a-1)x+2]/(x-1)=0
The fraction has no solution
① There is an increasing root, that is, x = 1
a-1+2=0
a=-1
② No solution
a-1=0
a=1
That's right

If the equation AX-5 / x + 4-2 = 0 about X has no solution, then the value of a is

Is this equation? (AX-5) / (x + 4) - 2 = 0ax-5 - 2x-8 = 0 (A-2) x = 13 when a = 2, no matter what value x is, the equation does not hold, that is, the equation has no solution. When a ≠ 2, x = 13 / (A-2) equation has no solution, that is, the solution is x = - 4, (- 4 is the increasing root of the equation), that is, 13 / (A-2) = - 4-4 (A-2) = 13-4a + 8 = 13A = - 5 / 4

The equation AX + 1 = B with respect to X has a unique solution when a and B are of any value,

Ax + 1 = B has unique solution
When a ≠ 0
ax=b-1
x=（b-1）/a
There are countless solutions
mean
When a = 0, B-1 = 0,
That is, a = 0, B = 1
unsolvable
When a = 0, B ≠ 1
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