How do you solve #x^2/(x^2-4) = x/(x+2)-2/(2-x)#?

Answer 1

I found no solutions (the left side is NOT equal to the right one)

Let us write it as: #x^2/(x^2-4)=x/(x+2)color(red)(+2/(x-2))# we can use as common denominator #(x^2-4)# which is equal to #(x+2)(x-2)# and write: #x^2/cancel((x^2-4))=(x(x-2)+2(x+2))/cancel((x^2-4))# giving: #cancel(x^2)=cancel(x^2)cancel(-2x)+cancel(2x)+4#
so we get #4=0# which is not true
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Answer 2

To solve the equation x^2/(x^2-4) = x/(x+2) - 2/(2-x), we can follow these steps:

  1. Start by finding a common denominator for the fractions on the right side of the equation. The common denominator is (x+2)(2-x).

  2. Rewrite the fractions on the right side with the common denominator: x/(x+2) = x(2-x)/[(x+2)(2-x)] -2/(2-x) = -2(x+2)/[(x+2)(2-x)]

  3. Simplify the fractions on the right side: x/(x+2) = (2x-x^2)/[(x+2)(2-x)] -2/(2-x) = -2(x+2)/[(x+2)(2-x)]

  4. Combine the fractions on the right side: (2x-x^2)/[(x+2)(2-x)] - 2(x+2)/[(x+2)(2-x)]

  5. Now, we can eliminate the denominators by multiplying both sides of the equation by [(x+2)(2-x)]: x^2 - 2x - x^2 - 4x - 4 = 2x - x^2 - 4x - 8

  6. Simplify the equation: -6x - 4 = -2x - 8

  7. Combine like terms: -6x + 2x = -8 + 4 -4x = -4

  8. Divide both sides by -4: x = 1

Therefore, the solution to the equation x^2/(x^2-4) = x/(x+2) - 2/(2-x) is x = 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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