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anonymous
 5 years ago
pls sove this by partial fraction method(x^2+1)/x(x1)(x+1)
anonymous
 5 years ago
pls sove this by partial fraction method(x^2+1)/x(x1)(x+1)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Start by expanding (x1)(x+1) to get x^21. Then try this anzats (or estimation): A/x + (Bx+c)/(x^21). You can then cross multiply to get: \[x ^{2}+1=A(x^{2}1)+x(Bx+C)\] Is that enough to work with? I don't know what you mean by "solve" in this context but if you try different x values in the above  or expand out the LHS  you can find A, B & C, which you just sub back in the anzats to find the partial fractions. I hope that helps :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have tried and got the\[(1\div x)+1\div(x+1)+1\div(x1)\] answer ...is this right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got a=1 b=1 and c=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You may be right but I found A by x=0 and B & C by considering x = 1 and x = 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and my text says 1/(x1)+1/(x+1)1/x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0o.k will try it once again ....thanks alot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the text book says A=1 ok

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok as long as you got the correct solution and answer....LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When x=0, 1=A(1) <=> A=1. When x=1, 2=A(11) +(B+C) and when x=1, 2=A(11) +(BC) <=> C=0 & B=2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mhhhmm thanks both of you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 0=Bx  Cx therefore 0=x(BC) B=C ..If B=1=C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where did you get 0=BxCx from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0expand x^2 + 1 = A(x1)(x+1) + Bx(X+1) +Cx(x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is incorrect: C is the constant term. There is no Cx.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0=Ax^2A + Bx^2+Bx+Cx^2Cx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mark the quation you have given might be correct ....can please elaborate for the value of A

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0WHAT VALUE OF "X" DID YOU TAKE TO FIND THE VALUE OF "A"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No; (x+1)/(x^21)  1/x does not work. But (2x)/(x^21)  1/x does.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from (x^2 +1) A B C  =  +  +  x(x1)(x+1) x x1 x+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was not using that anzats lol. rahulyadav was correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2 + 1 = A(x1)(x+1) + Bx(x+1) + Cx(x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mark o. to find value of A what value of x did you take???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When x=0, 1=A(1)(1) <=> A=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am not confident about the value of which i got cause the coefficient of A ALWAYS COMES DOWN TO "0" IF I PUT X=1 OR X=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0=A(x^2 1)+B(x^2 +x) +C(x^2x) x^2+1 = Ax^2  A + Bx^2 + Bx + Cx^2 Cx 1=A......A=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That does not matter if you have already found A using x=0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00=BxCx 0=x(BC) 0=(BC)... B=C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so finally both the values work for me but which value should i use?...x=1,x=1,or x=0????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The purpose of using x values is to discover what A, B & C are. The anzats mark put up is called an identity. It works for all x values.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks both of you for giving so much time to this problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just did it a long way.... and you can do it a short cut method like what shaun did....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok welcome and good luck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Strictly speaking, the "=" sign should have three lines to make that clear. When writing an anzats for partial fractions, try using the three lines to remind you that it holds for any x. And you're welcome :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes correct you can use 3 lines in equal sign
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