[3-(1/x)]/[9-(1/x^2)]
2007-06-06 12:48:16 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Multiply by x²/x²
x²[3-(1/x)]/(x²[9-(1/x^2)])
[3x²-x]/[9x²-1]
Now factor
x[3x-1]/[(3x-1)(3x+1)]
Divide both sides by common factors
x/(3x+1)
2007-06-06 12:55:53 · answer #1 · answered by Astral Walker 7 · 0⤊ 0⤋
lets break this up a little
3-(1/x) = (3x-1)/x
9- (1/x^2) = (9x^2 - 1)/X^2
divided those two and you get:
(3x-1)/[9x^2-1]/x
use division...
x(3x-1)/ ((3x-1)(3x+1)
answer = x/(3x+1)
2007-06-06 19:54:40 · answer #2 · answered by PurpleAndGold10 3 · 0⤊ 0⤋
[3-(1/x)]/[9-(1/x^2)]
Thanks for grouping your terms.
((3x - 1)/x)/((9x² - 1)x²)
invert and multiply
((3x - 1)/x)*(x²/(9x² - 1))
((3x - 1)/x) * x²/((3x - 1)(3x + 1))
simplify
x/(3x + 1)
.
2007-06-06 20:00:40 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
[3-(1/x)]/[9-(1/x^2)]
=(3-1/x)/(9-1/x^2)
={(3x-1)/x}/{(9x^2-1)/x^2}
=[(3x-1)/x}/{(3x+1)(3x-1)/x^2}
=(3x-1)/x *x^2/(3x+1)(3x-1)
=x/(3x+1)
2007-06-06 20:00:36 · answer #4 · answered by alpha 7 · 0⤊ 0⤋