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RolyPolyBest ResponseYou've already chosen the best response.0
\[\frac{x^2}{1x^2} = \frac{x^2}{(1x)(1+x)}\]How to continue??
 one year ago

nitzBest ResponseYou've already chosen the best response.0
\[x ^{2}/(1x)(1+x)=A/(1x)+B(1+x)\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
But that doesn't seems to work.. Here's what I mean: A(1+x) + B(1x) = x^2 (AB)x +(A+B) = x^2 => AB =0 , A+B =0 Not reasonable.
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
The degree of numerator must strictly be less than degree of denominator.
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
so, one thing u can do is, resolve 1/(1x^2) into partial fractions and then multiply x^2 to both terms.
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
\(\large \frac{(x^2)}{(1x^2) } = \frac{(1x^2)1}{1x^2} = 1\frac{1}{1x^2}\)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
A(1+x) + B(1x) = x^2 for x= 1 2 B = x^2 B = x^2/2 for x= 0 A + B = x^2 for x= 1 2A = x^2 A = x^2/2
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Hmmm... 1 1 = 2?! Shouldn't it be \[\frac{(x^2)}{(1x^2) } = \frac{(1x^2)+1}{1x^2} = 1+\frac{1}{1x^2}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
@UnkleRhaukus A(1+x) + B(1x) = x^2 for x= 1 2 B = (1)^2 B = 1/2 for x= 0 A + B = 0 for x= 1 2A = 1^2 A = 1/2 And it doesn't work here :S
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
@hartnn Yes! When can I be as smart as you???!!!!!
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
with just a little bit of practice :P
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
\[\frac{x^2}{1x^2}=x^2\times\frac{1}{1x^2} = x^2\times\frac{1}{(1x)(1+x)}\] \[\frac{1}{(1x)(1+x)}=\frac A{(1−x)}+\frac B{(1+x)}=\frac{A(1+x)+B(1x)}{(1x)(1+x)}\] \[1=A(1+x)+B(1x)\] for \(x=1\) \(1=2A\) \(A=1/2\) for \(x=1\) \(1=2B\) \(B=1/2\) \[\frac{1}{(1x)(1+x)}=\frac{1/2}{(1−x)}\frac {1/2}{(1+x)}\] \[\frac{x^2}{(1x)(1+x)}=\frac{x^2}{2(1−x)}\frac {x^2}{2(1+x)}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Excuse.. me.... When x =1 1 = A(1+(1)) + B(1(1)) 1 = 0 + 2B B = 1/2?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=Resolve+into+partial+fraction%3A+x%5E2+%2F+%281x%5E2%29 I don't know how to get that 1/(2(1x))
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Such simple maths is killing me :\ Thanks!!!!!!!!!
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
im not sure about the 1
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Here it goes~ \[\frac{x^2}{1x^2}\]\[=\frac{(x^2)}{1x^2}\]\[=\frac{(1x^2)+1}{1x^2}\]\[=1+\frac{1}{1x^2}\] \[\frac{1}{1x^2}=\frac{A}{1+x}+\frac{B}{1x}\]A(1x)+B(1+x)=1 A = B = 1/2 So, \[\frac{1}{1x^2}=\frac{1}{2(1+x)}+\frac{1}{2(1x)}\] Thus, \[\frac{x^2}{1x^2}=1+\frac{1}{2(1+x)}+\frac{1}{2(1x)}\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
A = B = 1/2 great work @RolyPoly
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
what u have done is correct.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
:'( Last but not least... \[\frac{x^2}{1x^2}=1+\frac{1}{2(1+x)}\frac{1}{2(x1)}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
It needs tons of thousands of millions of practice!! :'( Thanks again!!!
 one year ago
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