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

phi
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0is their any easy way instead of just taking common denominator

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

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

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

Jonask
 one year 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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