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Can I open the denom with a^2 - b^2 = (a+b)*(a-b) and then cancel out x-2 from num and denom? Is this the right approach?
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umang.lakhani
\(\frac{x-2}{x^2-4} =1 \) which of the following can x be?

A. -1 and 2
B. -2
C. -3 and -1
D. -1
E. -2, -1 and 2

Can I open the denom with a^2 - b^2 = (a+b)*(a-b) and then cancel out x-2 from num and denom? Is this the right approach?

Yes, that would be correct:

    \(\frac{x-2}{x^2-4} =1 \);

    \(\frac{x-2}{(x-2)(x+2)} =1 \);

We can reduce by x - 2 here because it cannot be 0:

    \(\frac{1}{x+2} =1 \);

    \(1 = x+2\);

    \(x=-1\)
.

Answer: D.
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