1 over t+1 minus 1 over t-1
2006-12-16 07:07:18 · 4 answers · asked by highrllr0079 2 in Science & Mathematics ➔ Mathematics
1/(t+1)-1/(t-1)=(t-1)/(t+1)(t-1)-(t+1)/(t+1)(t-1)=
(t-1-(t+1))/(t+1)(t-1)=
(t-1-t-1)/(t-1)^2
-2/(t-1)^2
2006-12-16 07:20:42 · answer #1 · answered by wayne 4 · 0⤊ 0⤋
Answer: -2 over (t^2-1)
To simplify, multiply the first term by (t-1)/(t-1) and the second term by (t+1)/(t+1) to give them the same denominators. Distribute and then combine like terms.
2006-12-16 15:19:14 · answer #2 · answered by Anonymous · 0⤊ 0⤋
1/(t+1) - 1/(t-1)
The numerator, 1-1, equals 0.
Once you distribute the negative sign in the denominator, (t-1) - (t+1) to become t-1-t-1, the new denominator is -2
0/-2 equals 0.
2006-12-16 15:16:42 · answer #3 · answered by Anonymous :) 5 · 0⤊ 0⤋
1 over t+1 minus 1 over t-1
1/(t+1) - 1/(t-1)
= (t-1)/[(t-1)(t+1) - (t+1)/[(t-1)(t+1)]
=(t-1 -t -1)/[(t-1)(t+1)]
=-2/(t^2-1)
2006-12-16 15:18:29 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋