关于方程(a+(a+…+(a+x)~(1/2))~(1/2))~(1/2)=x与(a+x)~(1/2)=x的同解问题,[1]文已圆满地解决了。关于方程(a-(a-…-(a-x)~(1/2))~(1/2))~(1/2)=x与(a-x)~(1/2)=x的同解问题,[1]文只是指出它们一般不同解,至于它们在什么条件下同解,[1]文未讲。如果弄清了在某种条件下(a-(a-…-(a-x)~(1/2))~(1/2))~(1/2)=x与(a-x)~(1/2)=x同解,那么在这种条件下解前面这个方程就是非常方便的事情了。这就促使我们去探讨(a-(a-…-(a-x)~(1/2))~(1/2))~(1/2)=与(a-x)~(1/2)=x同解的条 About the equation (a+(a+...+(a+x)~(1/2))~(1/2))~(1/2)=x and (a+x)~(1/2)=x The problem of the same solution, [1] has been successfully resolved. About the solution of the equation (a-(a-...-(ax)~(1/2))~(1/2))~(1/2)=x and (ax)~(1/2)=x Question, [1] only points out that they generally have different solutions. As for the conditions under which they resolve the same solution, [1] does not speak. If you have made it clear under certain conditions (a-(a-...-(ax)~(1/2))~(1/2))~(1/2)=x and (ax)~(1/ 2) = x is the same solution, then it is very convenient to solve this equation in this condition. This leads us to explore (a-(a-...-(ax)~(1/2))~(1/2))~(1/2)=(ax)~(1/2)=x The same solution