(x^2 - 2x + 1)/(5x^2 - 20x + 15)
2007-12-17 14:25:37 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
=(x - 1)(x - 1) / [5(x^2 - 4x + 3)]
=(x - 1)(x - 1) / [5(x - 3)(x - 1)]
=(x - 1) / [5(x - 3)]
2007-12-17 14:30:25 · answer #1 · answered by Jacob A 5 · 0⤊ 0⤋
(x^2-2x+1) = (x-1)(x-1)
(5x^2-20x+15) = 5(x^2-4x+3)=5( x-3)(x-1)
(x^2-2x+1)/5x^2-20x+15) = (x-1)(x-1)/(x-1)(x-3)5 = (x-1)/5(x-3)
2007-12-17 22:37:51 · answer #2 · answered by LucaPacioli1492 7 · 0⤊ 0⤋
(x^2 - 2x + 1)/(5x^2 - 20x + 15)
= [ x^2 - x - x + 1] / [ 5x^2 - 5x - 15x + 15 ]
= [(x-1)(x-1)] / [ 5x(x -1) -15 (x-1)]
= (x-1)^2 / (x-1) (5x-15)
= (x-1) / (5x-15)
= (x-1) / 5(x-3)
2007-12-17 22:41:22 · answer #3 · answered by Natali 2 · 0⤊ 0⤋
(x^2-2x+1) / (5x^2-20x+15
(x-1)(x-1) / 5(x^2-4x+3)
(x-1)(x-1) / 5(x-1)(x-3)
(x-1)^2/ 5(x-3)
2007-12-17 22:30:21 · answer #4 · answered by Grampedo 7 · 0⤊ 1⤋
x-1 / 5(x-3)
2007-12-17 22:36:59 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(x-2)(x+1)/5((x-3)(x-1)
2007-12-17 22:29:21 · answer #6 · answered by someone else 7 · 0⤊ 1⤋
(x-1)(x-1)/5(x-3)(x-1) X-1 cancel so you have X-1/5(x-3)
2007-12-17 22:31:05 · answer #7 · answered by swimmer12_9 2 · 0⤊ 0⤋
(x^2-2x+1)/(5x^2-20x+15)
(x-1)(x-1)/(5x-15) (x-1)
x-1/5x-15
x-1/ 5(x-3)
2007-12-17 22:28:55 · answer #8 · answered by Stephen Y 6 · 0⤊ 2⤋