2006-09-24 05:45:35 · 6 answers · asked by jairres t 1 in Science & Mathematics ➔ Mathematics
yep
so what is the question?
2006-09-24 05:47:23 · answer #1 · answered by Anonymous · 0⤊ 0⤋
2
2006-09-24 12:52:26 · answer #2 · answered by Jax 4 · 0⤊ 0⤋
I guess you are looking at factorisation:
(x^2+2xy+y^2)/(x^2-y^2) * (2x^2-xy-y^2)/(x^2-xy-2y^2)
[ (x+y)(x+y) ] / [ (x+y)(x-y) ] * (2x^2-xy-y^2)/(x^2-xy-2y^2)
[ (x+y) ] / [ (x-y) ] * (2x^2-xy-y^2)/(x^2-xy-2y^2)
[ (x+y)*(2x^2-xy-y^2) ] / [ (x-y)*(x^2-xy-2y^2) ]
[ (2x^3-x^2y-xy^2+2yx^2-xy^2-y^3) ] / [ (x^3-x^2y-2xy^2-yx^2+xy^2+2y^3) ]
[ (2x^3+yx^2-2xy^2-y^3) ] / [ (x^3-2yx^2-xy^2+2y^3) ]
let us now do the division
[ 2(x^3-2yx^2-xy^2+2y^3 +2yx^2+xy^2-2y^3+(1/2)yx^2-xy^2-(1/2)y^3) ] / [ (x^3-2yx^2-xy^2+2y^3) ]
2 + [ (4yx^2+2xy^2-4y^3+yx^2-2xy^2-y^3) ] / [ (x^3-2yx^2-xy^2+2y^3) ]
2 + [ (5yx^2-5y^3) ] / [ (x^3-2yx^2-xy^2+2y^3) ]
2 + [ 5y (x^2-y^2) ] / [ (x^3-2yx^2-xy^2+2y^3) ]
let us bring back the factorisation of the term below
2 + [ 5y (x-y)(x+y) ] / [ (x-y)*(x^2-xy-2y^2) ]
2 + [ 5y (x+y) ] / [ (x^2-xy-2y^2) ]
2006-09-24 12:59:37 · answer #3 · answered by sebourban 4 · 1⤊ 0⤋
x^2 +2xy -3y^2
2006-09-24 12:54:47 · answer #4 · answered by HeartBreakKid 2 · 0⤊ 0⤋
the answer will be a number sentence
2006-09-24 12:55:00 · answer #5 · answered by moo 3 · 0⤊ 0⤋
Uhh..why don't you do it??
2006-09-24 12:48:42 · answer #6 · answered by [brown♥eyed♥girl] 4 · 0⤊ 0⤋