{(n/3)-(3/n)} / {1+(n/3)}
Simply the complex fraction.
2007-02-23 12:03:09 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
{(n/3)-(3/n)} / {1+(n/3)}
= [(n^2 - 9)/3n] / ((3+n)/3)
= [(n^2 - 9)/3n] × (3/(3+n))
= [ (n+3)(n-3) × 3] / 3n(n+3)
= (n-3)/n = 1-3/n
2007-02-23 12:15:26 · answer #1 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋
multiply by 3n in denominator n numerator
(n^2-9)/(3n+n^2)=(n-3)(n+3)/n(n+3)=(n-3)/n=1-3/n
2007-02-23 12:18:44 · answer #2 · answered by ? 4 · 0⤊ 0⤋
= (n^2-9)/(3n+n^2) = (n+3)*(n-3)/n(n+3) =(n-3)/n
2007-02-23 12:10:31 · answer #3 · answered by santmann2002 7 · 1⤊ 0⤋
{(n/3)-(3/n)}/+{1+(n/3)}
=[(n^2-9)/3n}+{(3+n)/3}
=(n+3)(n-3)/3n }+{(n+3)/3}
={(n+3)(n-3)+n(n+3)}/3n
=(n^2-9+n^2+3n)/3n
=(2n^2+3n-9)/3n
=(2n-3)(n+3)/3n
2007-02-23 12:19:53 · answer #4 · answered by alpha 7 · 0⤊ 0⤋