Place in simplest form:
(x^2-9/3x^3) * (3x^2+6x/x^2+4x+3) / (x^2-4/3x+3)
Place "( )" as a fraction.
" ^ " is an exponet
2006-12-17 04:31:42 · 3 answers · asked by New York 1 in Science & Mathematics ➔ Mathematics
(x + 3)(x - 3)/3x³ * (3x(x + 2)/(x + 1)(x + 3)) * 3(x + 1)/(x + 2)(x - 2)
Cancelling:
(x - 3)/x² * 1 * 3/(x - 2)
Looks like that's as far as you can go, so:
3(x - 3) / [x²(x +2)]
2006-12-17 04:39:27 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
1+1 = 2
2006-12-17 12:39:11 · answer #2 · answered by gabnella 6 · 0⤊ 1⤋
x^2-9/3x^3=(x+3)(x-3)/3x^3
3x^2+6x/x^2+4x+3
=3x(x+2)/(x+3)(x+1)
x^2-4/3x+3=(x+2)(x-2)/3(x+1)
now plugging in the expression is
=(x+3)(x-3)(3x)(x+2)(3)(x+1)/
3x^3(x+3)(x+1)(x+2)(x-2)
=(3x/3x^3)(3)(x-3)/(x-2)
=9(x-3)/x^2(x-2)
2006-12-17 12:42:07 · answer #3 · answered by raj 7 · 0⤊ 0⤋