2007-09-11 21:44:45 · 4 answers · asked by platinumbutterfly713 1 in Science & Mathematics ➔ Mathematics
27 / (x^2 - 81) + 3 / (2(x+9))
Factorise x^2 - 81: it's a difference of perfect squares, so it factorises as (x-9) (x+9). So the lowest common denominator is 2 (x-9) (x+9). Put everything over this common denominator and we get
[27 (2)] / [2 (x-9) (x+9)] + [3 (x-9)] / [2 (x+9) (x-9)]
= [54 + 3(x-9)] / [2 (x-9) (x+9)]
= (3x + 27) / [2 (x-9) (x+9)]
= 3 (x+9) / [2 (x-9) (x+9)]
= 3 / [2 (x-9)]
2007-09-11 21:54:14 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
27/(x^2-81) + 3/2(x+9)
in the first term the denominator x^2-81 = (x+9)(x-9) since a^2-b^2 = (a+b) (a-b)
=27/(x+9)(x-9) + 3/2(x+9)
=2*27/2 *(x+9)(x-9) + 3(x-9)/2*(x+9)(x-9)
=54+3(x-9)/ 2(x-9)(x+9)
=54+3x-27 / 2(x-9)(x+9)
=3x+27 / 2(x-9)(x+9)
=3(x+9)/ 2(x-9)(x+9)
there is x+9 both in the numerator and denominator so it gets cancelled
=3/2(x-9)
2007-09-11 21:55:39 · answer #2 · answered by shubham_nath 3 · 0⤊ 0⤋
27 over x²- 81) plus (3 over 2(x+9))
=54/2(x-9)(x+9) + 3(x-9)/2(x-9)(x+9)
=(54+3x-27)/2(x-9)(x+9)
=3(x+9)/2(x-9)(x+9)
=3/2(x-9)
2007-09-11 21:53:07 · answer #3 · answered by ngtuhanh188 3 · 0⤊ 0⤋
1. X*-81=(x-9)(x+9) , 27/(x-9)(x+9)+3/ 2/(x+9)=54+3x-27/2(x-9)(x+9)=3x-27/2(x-9)(x+9)=3(x-9)/2(x-9)(x+9)=3/2(x+9)
2007-09-11 22:00:42 · answer #4 · answered by lbelfer 4 · 0⤊ 0⤋