Please add the following polynomial fractions.
(a - 1/a + 1)+(a + 1/ a - 1)=
does it equal a^4??
2007-07-18 10:33:17 · 9 answers · asked by soccermom71968 1 in Science & Mathematics ➔ Mathematics
or would it equal 1??
2007-07-18 10:37:47 · update #1
yup definatelly
2007-07-18 10:35:37 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Grompfet, Hustolemyname and Ironduke got it. Correcting a couple of typos in Ironduke's answer:
{(a-1)^2 +(a+1)^2}/(a^2-1) = 2(a^2+1)/(a^2-1)
2007-07-18 18:20:14 · answer #2 · answered by recordsetter01 2 · 0⤊ 0⤋
[ a - (1/a) + 1 ] + [a + (1/a) - 1]
= [a^2 - 1 + a + a^2 + 1 - a] / a
= [2a^2] / a
= 2 a
2007-07-18 17:45:29 · answer #3 · answered by Jeƒƒ Lebowski 6 · 0⤊ 0⤋
first find a common denominator which will be (a+1)(a-1) so multiply the first by (a-1)/(a-1) and the second by (a+1)(a+1) so you get:
[(a-1)^2+(a+1)^2] / [(a+1)(a-1)]
which simplifies to
2(a^2+1) / (a^2-1)
2007-07-18 17:38:54 · answer #4 · answered by grompfet 5 · 0⤊ 0⤋
Alright, first, find a common denominator, which will be (a+1)(a-1). In order to do this, you cross multiply.
(a-1) (a+1) (a-1)^2 + (a+1)^2
------ * (a-1) + -------- * (a+1).you end up with ---------------------
(a+1) (a-1) (a+1)(a-1)
In the end, you get a^2+2/(a+1)(a-1)
2007-07-18 17:46:18 · answer #5 · answered by rotcfreak1 5 · 0⤊ 0⤋
i think you mean
(a-1)/(a+1) + (a+1)/(a-1) =
[ (a-1)^2 + (a+1)^2 ] / [(a+1)(a-1) ] =
[ 2 a^2 + 2 ] / [ a^2 - 1 ] =
2 + 4/(a^2-1)
(which is same as 2(a^2+1)/(a^2-1) but I prefer my "shape" )
2007-07-18 17:39:17 · answer #6 · answered by hustolemyname 6 · 0⤊ 0⤋
= (a + a) + (1/a - 1/a) + (1 - 1)
= 2a
2007-07-21 04:23:47 · answer #7 · answered by Como 7 · 0⤊ 0⤋
No.
It is {(a-1)^2 +(a+1)^2/}[(a^2-1)] = 2(a^2+1)/(a^2-1)
2007-07-18 17:51:13 · answer #8 · answered by ironduke8159 7 · 0⤊ 0⤋
no
answer: 2a
2007-07-18 17:38:06 · answer #9 · answered by Robin 4 · 0⤊ 0⤋