\(\frac{2x^2 - 3x - 4}{(x-2)(x+3)}\)
\(\frac{2x^2 + 3x - 4}{(x-2)(x+3)}\)
\(\frac{2}{(x-2)(x+3)}\)
\(\frac{ 3x + 4}{(x-2)(x+3)}\)
\(\frac{3x - 4}{(x-2)(x+3)}\)
Correct answer is D
\(\frac{x}{x-2}-\frac{x+2}{x+3}\)
\(\frac{[x][x+3] - [x+2][x-2]}{[x-2][x+3]}\)
= \(\frac{x^2 + 3x - x^2 - 4}{[x-2][x+3]}\)
= \(\frac{3x - 4}{[x-2][x+3]}\)
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