Simplify!
[ (x + 1) / (x - 1) ] - [ (x - 1) / (x + 1) ]
*Ans: 4x / (x^2 - 1)*
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2006-11-01 16:00:57 · 11 answers · asked by Moo 4 in Science & Mathematics ➔ Mathematics
[(x+1)/(x-1)]-[(x-1)/(x+1)]
Finding LCM of denominator (x-1) and (x+1) gives (x-1)(x+1)
Now making it a like fraction will give,
=[(x+1)^2 - (x-1)^2] / (x-1)(x+1)
we have (x-1).(x+1)=x^2-1
also if you expand the numerator and cancell the x^2 and constant 1^2terms
the answer is
=4x/(x^2-1)
2006-11-01 16:07:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
3
2006-11-01 16:07:52 · answer #2 · answered by Anonymous · 0⤊ 0⤋
3
2006-11-01 16:03:50 · answer #3 · answered by stephaniencurtis 2 · 0⤊ 0⤋
Find common denominator = (x+1)*(x-1) = (x^2 - 1)
The fraction becomes {(x+1)^2 - (x-1)^2}/(x^2-1)
Multiply the numerator out
{x^2 + 2x + 1) - (x^2 - 2x +1) = 4x
The result is then 4x/(x^2-1)
2006-11-01 16:08:22 · answer #4 · answered by gp4rts 7 · 0⤊ 0⤋
Get common denominator:
[(x + 1)^2 - (x - 1)^2] / [(x - 1)(x + 1)]
= (x^2 + 2x + 1 - x^2 + 2x - 1) / (x^2 - 1)
= 4x / (x^2 - 1)
2006-11-01 16:04:51 · answer #5 · answered by jacinablackbox 4 · 0⤊ 0⤋
a million. 30 cm 2. 4.2 m 3. 250 mL 4. 8,3 hundred mm 5. 6 km 6. 80 5 mg 7. 71.24 kg 8. seven-hundred m 9. 4.8 L 10. 4,800 cm I guessed on like multiple of those different than 4 thte 1st one! with somewhat of luck I relatively have been given them precise!!
2016-11-26 23:05:04 · answer #6 · answered by ? 3 · 0⤊ 0⤋
(x+1)(x+1)-(x-1)(x-1) / (x+1)(x-1)
((x^2 +2x + 1) - (x^2 -2x +1)) / (x^2 -1)
x^2 cancels, as does 1. 2x - (-2x) = 4x
(4x)/(x^2 -1)
2006-11-01 16:07:20 · answer #7 · answered by xenrous 2 · 0⤊ 0⤋
[(x+1) / (x-1)] - [(x-1) / (x+1)]
= [(x+1)(x+1) - (x-1)(x-1)] / [(x-1)(x+1)]
= [x^2 +2x +1 - (x^2 - 2x +1)] / [x^2 - 1]
= [4x / (x^2 - 1)]
2006-11-01 16:07:05 · answer #8 · answered by jayde 2 · 0⤊ 0⤋
[ (x + 1) / (x - 1) ] - [ (x - 1) / (x + 1) ]
[ (x^2 +2x+ 1) /(x-1) (x+1)] - [ (x^2 - 2x+ 1)/ (x +1) (x-1)]
(x^2 +2x+1 - x^2 + 2x -1)/ (x+1)(x-1) =
4x/(x^2-1)
2006-11-01 16:08:51 · answer #9 · answered by Sheigh L 2 · 0⤊ 0⤋
of course 0
2006-11-01 16:03:51 · answer #10 · answered by roman 3 · 0⤊ 0⤋