1. \(\displaystyle \dfrac{x^{-2}-a^{-2}}{x-a}\)
At first, I was totally confuse as to what to do. I always picture the negative exponents as putting the terms under the bar, and making them part of the denominator. But both terms in the numerator have negative exponents and that would seem to mean that nothing at all should be in the numerator position, but that doesn't make sense.
I then figured out that I should make them their own fractions above the line:
\(\displaystyle \dfrac{\frac{1}{x^2}-\frac{1}{a^2}}{x-a}\)
Then I just multiplied the quotients in the numerator by the alternative denominators (over those same denominators, to make it 1) to get a common denominator in the numerator position:
\(\displaystyle \dfrac{\frac{1}{x^2}\cdot{\frac{a^2}{a^2}}-\frac{1}{a^2}\cdot{\frac{x^2}{x^2}}}{x-a}\)
Alright, so from here I ended up just taking the denominator term, flip it, and make it something to multiply the other terms by.
\(\displaystyle (\dfrac{a^2}{a^2x^2}-\dfrac{x^2}{a^2x^2})\cdot\dfrac{1}{x-a}=\dfrac{a^2-x^2}{a^2x^2(x-a)}\)
So I then converted the binomial in the new numerator, and then stepped back divided the denominator \(\displaystyle x-a\) by -1
\(\displaystyle \dfrac{(a+x)(a-x)}{(a^2x^2)-1(a-x)}\)
From there I factor/canceled out the \(\displaystyle (a-x)\) to make:
\(\displaystyle \dfrac{a+x}{-a^2x^2}\cdot\dfrac{-1}{1}=\dfrac{-a-x}{a^2x^2}\)
At that point I thought that I was done, but I ended up with an answer that was different from the one the book gave. So my question is was there a way a might've directly ended up with the answer that the book gave? Here it was:
\(\displaystyle \dfrac{-x-a}{a^2x^2}\)
At first, I was totally confuse as to what to do. I always picture the negative exponents as putting the terms under the bar, and making them part of the denominator. But both terms in the numerator have negative exponents and that would seem to mean that nothing at all should be in the numerator position, but that doesn't make sense.
I then figured out that I should make them their own fractions above the line:
\(\displaystyle \dfrac{\frac{1}{x^2}-\frac{1}{a^2}}{x-a}\)
Then I just multiplied the quotients in the numerator by the alternative denominators (over those same denominators, to make it 1) to get a common denominator in the numerator position:
\(\displaystyle \dfrac{\frac{1}{x^2}\cdot{\frac{a^2}{a^2}}-\frac{1}{a^2}\cdot{\frac{x^2}{x^2}}}{x-a}\)
Alright, so from here I ended up just taking the denominator term, flip it, and make it something to multiply the other terms by.
\(\displaystyle (\dfrac{a^2}{a^2x^2}-\dfrac{x^2}{a^2x^2})\cdot\dfrac{1}{x-a}=\dfrac{a^2-x^2}{a^2x^2(x-a)}\)
So I then converted the binomial in the new numerator, and then stepped back divided the denominator \(\displaystyle x-a\) by -1
\(\displaystyle \dfrac{(a+x)(a-x)}{(a^2x^2)-1(a-x)}\)
From there I factor/canceled out the \(\displaystyle (a-x)\) to make:
\(\displaystyle \dfrac{a+x}{-a^2x^2}\cdot\dfrac{-1}{1}=\dfrac{-a-x}{a^2x^2}\)
At that point I thought that I was done, but I ended up with an answer that was different from the one the book gave. So my question is was there a way a might've directly ended up with the answer that the book gave? Here it was:
\(\displaystyle \dfrac{-x-a}{a^2x^2}\)