If the fractional equation (x-a) / (x-1) - 3 / x = 1 has no solution, then a= Double x (x-1) x(x-a)-3(x-1)=x(x-1) x²-ax-3x+3=x²-x (a+2)x=3 If a = - 2 Then a + 2 = 0 Then 0 = 3, not true, If a ≠ - 2 Then x = 3 / (a + 2), if there is no solution, then this is an increasing root That is, the common denominator is 0 x(x-1)=0 x=0,x=1 3 / (a + 2) = 0 doesn't hold. Why can't x = 0 be brought into the fractional equation? Then a is an arbitrary value!

If the fractional equation (x-a) / (x-1) - 3 / x = 1 has no solution, then a= Double x (x-1) x(x-a)-3(x-1)=x(x-1) x²-ax-3x+3=x²-x (a+2)x=3 If a = - 2 Then a + 2 = 0 Then 0 = 3, not true, If a ≠ - 2 Then x = 3 / (a + 2), if there is no solution, then this is an increasing root That is, the common denominator is 0 x(x-1)=0 x=0,x=1 3 / (a + 2) = 0 doesn't hold. Why can't x = 0 be brought into the fractional equation? Then a is an arbitrary value!

First of all, before you multiply x (x-1), you should assume that x ≠ 1 and 0, and then solve the equation, and then get (a + 2) x = 3. When a = - 2, there is no solution, and if the solution is 1 or 0, it does not conform to the assumption, and there is no solution. You can substitute x = 1 or 0 into the final equation, and get a = 1 when x = 1