help i rlly need fast answers thanks very much

The equation simplifies to:

\((x-1)(x-2)(x-3) \dots (x-100) = 0\)

There are 100 solutions, namely x = 1 to x = 100.

I don't think it does.

I'm pretty sure the constant is squared and not the whole thing.

So like \((x-1)(x-4)(x-9)...(x-10000)\)

Note  that  x  cannot   be   any perfect  square  from  x =  1  to  x = 100  (i.e.,  x= 1^2  to x = 10^2)  because  any  of  these  make  the  denominator  = 0

So......we  only  have   90   possible   x values  for  the  numerator