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\[y^4+\frac{ 1 }{ 16y^4 }+\frac{ 1 }{ 2 }\]
perfect square

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Other answers:

  • phi
you could make 16y^4 the common denominator and add up the terms the denominator is a perfect square. Does the numerator factor?
is their any easy way instead of just taking common denominator
i know a^2+2ab+b^2
  • phi
It is not that hard
  • phi
16 y^8 + 1 + 8 y^4
  • phi
re-arrange as 16 y^8 + 8 y^4 + 1 if you let y^4 = x, this is also 16x^2 + 8x +1
i just tried\[(y^2+\frac{1}{4y^2})^2\]
  • phi
if it is square your only hope is (4x+1)^2 which works
is that =0 because what happened to common denominator
is it not true to say\[(y^2+\frac{1}{4y^2})^2=y^4+2(y^2)(\frac{1}{4y^2})+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}\]

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