precal help

**Kingreaper** · September 15th 2008, 03:30 PM

simplify: (x-(1/x^2))/(x-(1/x^3))

simplify: 4+___d___
4+_4_
4+d

**icemanfan** · September 15th 2008, 03:35 PM

I'm not quite sure what the second question says, but for the first question you'll be needing common denominators. Working on the numerator first, we have $x - \frac{1}{x^2}$ . To get a common denominator, multiply $x \cdot \frac{x^2}{x^2}$ so that you get $\frac{x^3}{x^2} - \frac{1}{x^2}$ . Subtracting yields $\frac{x^3 - 1}{x^2}$ . Can you simplify the denominator?

**Kingreaper** · September 15th 2008, 03:46 PM

the bottom one, with many parantheses, says 4+(d/(4+(4/(4+d))))
i simplified the bottom part, but what do i do next?

**icemanfan** · September 15th 2008, 03:52 PM

For the first question, you should have gotten $\frac{x^4 - 1}{x^3}$ as the denominator, which leaves you with $\frac{\frac{x^3 - 1}{x^2}}{\frac{x^4 - 1}{x^3}}$ . Now, you can rearrange this (using the reciprocal rule) to $\frac{x^3(x^3 - 1)}{x^2(x^4 - 1)}$ . You can cancel out x^2 in the top and bottom to yield $\frac{x(x^3 - 1)}{x^4 - 1}$ and then remove a factor of x-1 from both top and bottom to yield $\frac{x(x^2 + x + 1)}{x^3 + x^2 + x + 1}$ and that's as simplified as it gets.

For the second question, using the parentheses, work from the inside out, using common denominators to combine fractions.

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