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x^2/(x-1)(x+1) = x^2/x^2-1 = 1,-1
Therefore,
x^2/(x-1)(x+1) = 1 + -1/(x-1)(x+1)
-1/(x-1)(x+1) = A/(x-1) + B/(x+1)
-1 = Ax + A + Bx - B
-1 = (A+B)x +A - B
A+B = 0, B = -A
-1 = A - B = A - (-A) = 2A
-1 = 2A, A = -1/2,
B = -(-1/2) = 1/2
Therefore,
x^2/(x-1)(x+1) = 1 - 1/2(x-1) + 1/2(x+1)
But somehow the answer given to me is
x^2/(x-1)(x+1) = 1 + 1/2(x-1) - 1/2(x+1)
Not sure which step I doing wrong, please help me on this. Thank you.
Hello, gilagila!
Quote:
$\displaystyle \frac{x^2}{(x-1)(x+1)}$
We have: .$\displaystyle \frac{x^2}{x^2-1} \;=\;\frac{x^2 - 1 + 1}{x^2-1} \;=\;\frac{x^2-1}{x^2-1} + \frac{1}{x^2-1} \;=\;1 + \frac{1}{x^2-1}$
Then: .$\displaystyle \frac{1}{(x-1)(x+1)} \;=\;\frac{A}{x-1} + \frac{B}{x+1}$
. . . . . . $\displaystyle 1 \;=\;A(x+1) + B(x-1)$
Let $\displaystyle x = 1\!:\;\;1 \;=\;A(2) + B(0) \quad\Rightarrow\quad A \,=\,\tfrac{1}{2}$
Let $\displaystyle x = \text{-}1\!:\;\;1 \;=\;A(0) + B(\text{-}2) \quad\Rightarrow\quad B \,=\,\text{-}\tfrac{1}{2}$
Therefore: .$\displaystyle \frac{x^2}{(x-1)(x+1)} \;=\; 1 + \frac{\frac{1}{2}}{x-1} - \frac{\frac{1}{2}}{x+1}$
Just want to check, how you change to this ??Quote:
Originally Posted by Soroban
$\displaystyle \frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}$Quote:
Originally Posted by gilagila
He subtracted 1 from the numerator and at the same time added it. This does not make any "change" to expression.
http://latex.codecogs.com/gif.latex?x^2$\displaystyle =$http://latex.codecogs.com/gif.latex?x^2http://latex.codecogs.com/gif.latex?...pace;-1+1
oic, thanks for explanation
