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Here is an example: $\frac{\nu y_0}{\left(x_0+\frac{1}{2}\right) \sqrt{\nu ^2+\frac{\left(4 x_0^2+4 y_0^2-1\right){}^2}{16 y_0^2}}}$

 test=(\[Nu] Subscript[y, 0])/((1/2+Subscript[x, 0]) Sqrt[\[Nu]^2+(-1+4 Subsuperscript[x, 0, 2]+4 Subsuperscript[y, 0, 2])^2/(16 Subsuperscript[y, 0, 2])])

I'd like to get rid of the denominator in the squareroot. I tried separating the numerator and denominator, and multiply both by $4y_0$ or $\sqrt{16y_0^2}$. Somehow neither works. Strangely, even if I multiply the denominator by $\sqrt{16y_0^2}$, the two terms won't combine together even if I use Simplify.

Currently I use FullSimplify[test,Subscript[y, 0]>0], which does not work. Hence, I had to copy the material in the squareroot, multiply by $16y_0^2$, simplify it separately, then plug back the original expression to obtain the results (the numerator will be multiplied by $4y_0$, too).

I wonder if there is any better way to do this mor efficiently?

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  • $\begingroup$ How about using a rule test /. {1/(16 Subsuperscript[y,0,2]) -> 1} $\endgroup$
    – user2757771
    Dec 17, 2019 at 16:26
  • $\begingroup$ This also might be a bit more generic test /. {1/ Sqrt[x_ + a_*y_/z_] -> 1/Sqrt[x + y]} $\endgroup$
    – user2757771
    Dec 17, 2019 at 16:30
  • $\begingroup$ A part of your problem is that you use Subsuperscript[y, 0, 2] instead of Subscript[y, 0]^2. Superscript is not a power, it is to make an upper index. You can make sure of it looking at FullForm[test]. $\endgroup$
    – Alexei Boulbitch
    Dec 17, 2019 at 16:47

2 Answers 2

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Try this:

test = (ν y)/((1/2 + 
      x) Sqrt[ν^2 + (-1 + 4 x^2 + 4 y^2)^2/(16 y^2)]);

test2 = MapAt[Together[#] &, test, {4, 1}]
test3 = MapAt[Hold, test2, {5, 1, 2}] // Simplify[#, y > 0] & //ReleaseHold

enter image description here

Have fun!

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  • $\begingroup$ Sorry, @Alexei Boulbitch, but this result is not true for negative y. % /. {x -> 3, y -> -2, \[Nu] -> 4} yields 128/(35 Sqrt[145]) whereas test /. {x -> 3, y -> -2, \[Nu] -> 4} yields -(128/(35 Sqrt[145])) . $\endgroup$
    – Akku14
    Dec 18, 2019 at 6:45
  • $\begingroup$ Have you noticed that I simplified with the condition y>0. For negative y values one simplifies with another condition. Generally, I only show the way of operation, not the exhaustive solution. $\endgroup$
    – Alexei Boulbitch
    Dec 18, 2019 at 9:56
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Or:

test = (\[Nu] y)/((1/2 + 
   x) Sqrt[\[Nu]^2 + (-1 + 4 x^2 + 4 y^2)^2/(16 y^2)]);

test // Together // 
        PowerExpand[#, Assumptions -> y \[Element] Reals && y != 0] &

enter image description here

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