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ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0\[y^4+\frac{ 1 }{ 16y^4 }+\frac{ 1 }{ 2 }\]

phi
 2 years ago
Best ResponseYou've already chosen the best response.0you could make 16y^4 the common denominator and add up the terms the denominator is a perfect square. Does the numerator factor?

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0is their any easy way instead of just taking common denominator

phi
 2 years ago
Best ResponseYou've already chosen the best response.0rearrange as 16 y^8 + 8 y^4 + 1 if you let y^4 = x, this is also 16x^2 + 8x +1

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.1i just tried\[(y^2+\frac{1}{4y^2})^2\]

phi
 2 years ago
Best ResponseYou've already chosen the best response.0if it is square your only hope is (4x+1)^2 which works

ksaimouli
 2 years ago
Best ResponseYou've already chosen the best response.0is that =0 because what happened to common denominator

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.1is 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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