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Simplify `(4x^2-x^(-2))/(2^0x^2-2^(-1))`

First note that `2^0=1` , so this term will disappear as multiplying by 1 does not change anything.

We can write any term with a negative exponent without a negative exponent by taking the reciprocal -- e.g. `a^(-1)=1/a` .

Thus `(4x^2-x^(-2))/(2^0x^2-2^(-1))=(4x^2-1/x^2)/(x^2-1/2)`

Now we can simplify the complex fraction by multiplying by ` ` `(2x^2)/(2x^2)` (This is the least common multiple of `x^2` and 2)

Now `(4x^2-1/(x^2))/(x^2-1/2)*(2x^2)/(2x^2)=(8x^4-2)/(2x^4-x^2)=(2(4x^4-1))/(x^2(2x^2-1))=(2(2x^2+1)(2x^2-1))/(x^2(2x^2-1))=(2(2x^2+1))/x^2`

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**This can be written as `(2(2x^2+1))/x^2=(4x^2+2)/x^2=4+2/x^2` ; any of these three would be correct.**

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** If you meant `4x^2-x^(-2)/(2^0x^2)-2^(-1)` you would get `4x^2-1/x^4-1/2` **

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