What is 1+(x^2-1/(4x^2))^2 ?

The solution to 1+(x^2-1/(4x^2))^2 is 1+((4x^4-1)^2)/(16x^4)

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simplify 1+(x^2-1/(4x^2))^2

simplify

1 + (x^{2} - \frac{1}{4 x ^{2}})^{2}

1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{16 x ^{4}}

Solution steps

1 + (x^{2} - \frac{1}{4 x ^{2}})^{2}

Join

x^{2} - \frac{1}{4 x ^{2}} : \frac{4 x ^{4} - 1}{4 x ^{2}}

= 1 + (\frac{4 x ^{4} - 1}{4 x ^{2}})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

(\frac{4 x ^{4} - 1}{4 x ^{2}})^{2} = \frac{( 4 x ^{4} - 1 ) ^{2}}{( 4 x ^{2} ) ^{2}}

= 1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{( 4 x ^{2} ) ^{2}}

Simplify

(4 x^{2})^{2} : 16 x^{4}

= 1 + \frac{( 4 x ^{4} - 1 ) ^{2}}{16 x ^{4}}

(2 x + 3) \cdot (2 x - 3)

x^{2} + 2 x + \frac{3}{4}

(3 + 4)^{2}

11 1^{2}

\frac{6 x ( x + 4 )}{x - 3} \cdot \frac{( x - 3 ) ( x - 6 )}{6 x ( x - 6 )}

What is 1+(x^2-1/(4x^2))^2 ?
The solution to 1+(x^2-1/(4x^2))^2 is 1+((4x^4-1)^2)/(16x^4)