What do you mean? That question doesn't make sense.
Answer:
We have the equation
![a+x= 1+xa^2](https://tex.z-dn.net/?f=a%2Bx%3D%201%2Bxa%5E2)
we leave the terms with x on the left side of the equation and the independent terms on the right side.
![x-xa^2=1-a\\(1-a^2)x=1-a\\](https://tex.z-dn.net/?f=x-xa%5E2%3D1-a%5C%5C%281-a%5E2%29x%3D1-a%5C%5C)
resolving for x we have that
![x=\frac{1-a}{1-a^2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1-a%7D%7B1-a%5E2%7D)
Since we need that the equation doesn't have solution, then it is necessary that the denominator of x be 0 and this occur when ![1-a^2=0\\1=a^2\\a=\pm1](https://tex.z-dn.net/?f=1-a%5E2%3D0%5C%5C1%3Da%5E2%5C%5Ca%3D%5Cpm1)
Then, for
the equation hasn't solution
Answer 3/5,1/5,1/6
:)))))))))