In the rational and radical expressions chapter of the book I'm going through I'm stumped as to how to make progress on these problems. The instructions are to, "Write as a single fraction in lowest terms. Do not rationalize denominators." Here's the first example:
\(\displaystyle \sqrt{x-2} +\frac{2}{\sqrt{x-2}}\)
With this I rationalized the numerator by multiplying the first side by \(\displaystyle \frac{\sqrt{x-2}}{\sqrt{x-2}}\) so I could have common denominators (even if not rational ones). This gave me
\(\displaystyle \frac{\sqrt{x-2}\cdot\sqrt{x-2}}{\sqrt{x-2}} +\frac{2}{\sqrt{x-2}}=\frac{x-2+2}{\sqrt{x-2}}=\frac{x}{\sqrt{x-2}}\)
That first problem made enough sense to me. But the subsequent problems tripped me up. The next example:
\(\displaystyle \frac{\sqrt{x^2-1}-\frac{x^2}{\sqrt{x^2-1}}}{x^2-1}\)
I was stuck at that point but the book said that I should multiply the whole thing (numerator and denominator) by the only internal denominator, \(\displaystyle \sqrt{x^2-1}\) which looks like
\(\displaystyle \frac{\sqrt{x^2-1}-\frac{x^2}{\sqrt{x^2-1}}}{x^2-1}\cdot\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}=\frac{x^2-1-x^2}{(x^2-1)\sqrt{x^2-1}}\)
I'm clear up to this point, but I don't understand the book when it writes the conclusion of problem. I suppose it's because I wasn't sure what
\(\displaystyle (x^2-1)\sqrt{x^2-1}\)
was supposed to equal. The books's final form for this was
\(\displaystyle -\frac{1}{(x^2-1)^{3/2}}\)
I can see how the numerator comes to negative 1, but the denominator eludes me. Help?
\(\displaystyle \sqrt{x-2} +\frac{2}{\sqrt{x-2}}\)
With this I rationalized the numerator by multiplying the first side by \(\displaystyle \frac{\sqrt{x-2}}{\sqrt{x-2}}\) so I could have common denominators (even if not rational ones). This gave me
\(\displaystyle \frac{\sqrt{x-2}\cdot\sqrt{x-2}}{\sqrt{x-2}} +\frac{2}{\sqrt{x-2}}=\frac{x-2+2}{\sqrt{x-2}}=\frac{x}{\sqrt{x-2}}\)
That first problem made enough sense to me. But the subsequent problems tripped me up. The next example:
\(\displaystyle \frac{\sqrt{x^2-1}-\frac{x^2}{\sqrt{x^2-1}}}{x^2-1}\)
I was stuck at that point but the book said that I should multiply the whole thing (numerator and denominator) by the only internal denominator, \(\displaystyle \sqrt{x^2-1}\) which looks like
\(\displaystyle \frac{\sqrt{x^2-1}-\frac{x^2}{\sqrt{x^2-1}}}{x^2-1}\cdot\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}=\frac{x^2-1-x^2}{(x^2-1)\sqrt{x^2-1}}\)
I'm clear up to this point, but I don't understand the book when it writes the conclusion of problem. I suppose it's because I wasn't sure what
\(\displaystyle (x^2-1)\sqrt{x^2-1}\)
was supposed to equal. The books's final form for this was
\(\displaystyle -\frac{1}{(x^2-1)^{3/2}}\)
I can see how the numerator comes to negative 1, but the denominator eludes me. Help?