Question

y^4+(1/16y^4)+1/2

Answer 1

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

Answer 2

perfect square

Answer 3

@Jonask

Answer 4

you could make 16y^4 the common denominator and add up the terms

the denominator is a perfect square. Does the numerator factor?

Answer 5

is their any easy way instead of just taking common denominator

Answer 6

@zepdrix

Answer 7

i know a^2+2ab+b^2

Answer 8

It is not that hard

Answer 9

16 y^8 + 1 + 8 y^4

Answer 10

=(a+b)^2

Answer 11

re-arrange as
16 y^8 + 8 y^4 + 1

if you let y^4 = x, this is also
16x^2 + 8x +1

Answer 12

i just tried\[(y^2+\frac{1}{4y^2})^2\]

Answer 13

if it is square your only hope is (4x+1)^2 which works

Answer 14

is that =0 because what happened to common denominator

Answer 15

(4x^4+1)^2

Answer 16

is it not true to say\[(y^2+\frac{1}{4y^2})^2=y^4+2(y^2)(\frac{1}{4y^2})+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}\]