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Old 2020-10-22, 11:37   #6
Dr Sardonicus
 
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Feb 2017
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Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives

2x^{\frac{1}{4}}\cdot\(x^{\frac{1}{2}}\;+\;x^{\frac{-1}{2}}\)^{\frac{1}{2}}

The "obvious" multiplication then gives 2\sqrt{x+1}.

I note that things can go wrong for complex values of "x" that aren't positive real numbers.
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