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anonymous

  • 5 years ago

pls sove this by partial fraction method(x^2+1)/x(x-1)(x+1)

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  1. anonymous
    • 5 years ago
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    Start by expanding (x-1)(x+1) to get x^2-1. Then try this anzats (or estimation): A/x + (Bx+c)/(x^2-1). 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 :)

  2. anonymous
    • 5 years ago
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    Here A= -1, B=2, C=0

  3. anonymous
    • 5 years ago
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    i have tried and got the\[(1\div x)+1\div(x+1)+1\div(x-1)\] answer ...is this right

  4. anonymous
    • 5 years ago
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    i got a=1 b=1 and c=1

  5. anonymous
    • 5 years ago
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    You may be right but I found A by x=0 and B & C by considering x = 1 and x = -1.

  6. anonymous
    • 5 years ago
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    and my text says 1/(x-1)+1/(x+1)-1/x

  7. anonymous
    • 5 years ago
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    We are both correct.

  8. anonymous
    • 5 years ago
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    o.k will try it once again ....thanks alot

  9. anonymous
    • 5 years ago
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    i think A=-1,,B=C=1

  10. anonymous
    • 5 years ago
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    so the text book says A=-1 ok

  11. anonymous
    • 5 years ago
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    i mean A=1... ok

  12. anonymous
    • 5 years ago
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    no A= -1

  13. anonymous
    • 5 years ago
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    B=C=1

  14. anonymous
    • 5 years ago
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    yeah A=-1

  15. anonymous
    • 5 years ago
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    ok as long as you got the correct solution and answer....LOL

  16. anonymous
    • 5 years ago
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    When x=0, 1=A(-1) <=> A=-1. When x=1, 2=A(1-1) +(B+C) and when x=-1, 2=A(1-1) +(B-C) <=> C=0 & B=2.

  17. anonymous
    • 5 years ago
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    mhhhmm thanks both of you

  18. anonymous
    • 5 years ago
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    but 0=Bx - Cx therefore 0=x(B-C) B=C ..If B=1=C

  19. anonymous
    • 5 years ago
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    Where did you get 0=Bx-Cx from?

  20. anonymous
    • 5 years ago
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    expand x^2 + 1 = A(x-1)(x+1) + Bx(X+1) +Cx(x-1)

  21. anonymous
    • 5 years ago
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    That is incorrect: C is the constant term. There is no Cx.

  22. anonymous
    • 5 years ago
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    =Ax^2-A + Bx^2+Bx+Cx^2-Cx

  23. anonymous
    • 5 years ago
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    mark the quation you have given might be correct ....can please elaborate for the value of A

  24. anonymous
    • 5 years ago
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    WHAT VALUE OF "X" DID YOU TAKE TO FIND THE VALUE OF "A"

  25. anonymous
    • 5 years ago
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    No; (x+1)/(x^2-1) - 1/x does not work. But (2x)/(x^2-1) - 1/x does.

  26. anonymous
    • 5 years ago
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    from (x^2 +1) A B C ------- = --- + --- + ---- x(x-1)(x+1) x x-1 x+1

  27. anonymous
    • 5 years ago
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    I was not using that anzats lol. rahulyadav was correct.

  28. anonymous
    • 5 years ago
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    x^2 + 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)

  29. anonymous
    • 5 years ago
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    Here A=-1 and B=C=1.

  30. anonymous
    • 5 years ago
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    mark o. to find value of A what value of x did you take???

  31. anonymous
    • 5 years ago
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    Try x=0.

  32. anonymous
    • 5 years ago
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    When x=0, 1=A(-1)(1) <=> A=-1.

  33. anonymous
    • 5 years ago
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    i 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

  34. anonymous
    • 5 years ago
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    =A(x^2 -1)+B(x^2 +x) +C(x^2-x) x^2+1 = Ax^2 - A + Bx^2 + Bx + Cx^2 -Cx 1=-A......A=-1

  35. anonymous
    • 5 years ago
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    That does not matter if you have already found A using x=0.

  36. anonymous
    • 5 years ago
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    0=Bx-Cx 0=x(B-C) 0=(B-C)... B=C

  37. anonymous
    • 5 years ago
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    so finally both the values work for me but which value should i use?...x=1,x=-1,or x=0????

  38. anonymous
    • 5 years ago
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    All three.

  39. anonymous
    • 5 years ago
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    yes it works also

  40. anonymous
    • 5 years ago
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    The 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.

  41. anonymous
    • 5 years ago
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    thanks both of you for giving so much time to this problem

  42. anonymous
    • 5 years ago
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    i just did it a long way.... and you can do it a short cut method like what shaun did....

  43. anonymous
    • 5 years ago
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    ok welcome and good luck

  44. anonymous
    • 5 years ago
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    Strictly 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 :)

  45. anonymous
    • 5 years ago
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    yes correct you can use 3 lines in equal sign

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