Question 118813
How to solve these 4 equations :
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These are not equations (no equal sign), you can't solve them;.
You can factor/simplify them, which is probably what you want you to do:
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1). {{{48ab^2c(4abc^0)^-2}}}
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First of all any number, raised to ^0 = 1 so we can write:
{{{48ab^2c(4ab)^-2}}}
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A negative exponent can be made positive by using the reciprocal, so we have:
{{{(48ab^2c)/(4ab)^2}}}
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Square each term inside the brackets (denominator)
{{{(48ab^2c)/(16a^2b^2)}}}
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CAncel and you have:
{{{(3c)/a}}}
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2). {{{(3x^0-n^-2)/(-(3m-n))}}}:
Follow the rules that we illustrated in the 1st one:
The reciprocal gets rid of the neg exponent, removing the bracket changes the signs
{{{(3 - (1/n^2))/(-3m+n)}}}; 
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Put the numerator into a single fraction form
{{{((3n^2 - 1)/n^2)/(-3m+n)}}}
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Which is:
{{{((3n^2 - 1)/n^2)}}} * {{{1/(-3m+n)}}} = {{{(3n^2-1)/(n^2(-3m+n))}}} = {{{(3n^2-1)/((-3mn^2+n^3))}}}
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3). {{{((144*125)/(2^3*3^2))^-1}}}
Do the math and invert to get rid of the ^-1:
{{{((144*125)/(8*9))^-1}}} = {{{(18000/72)^-1}}} = {{{72/18000}}} = {{{1/250}}}
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4). {{{(ab^-2) - (a^-2b)}}} = {{{a/b^2}}} - {{{b/a^2}}}; using the reciprocals to get rid of the negative exponents
To put this over a single denominator:
{{{(a^3 - b^3)/(a^2b^2)}}}