a, $\frac{4x + 1}{2}$ – $\frac{3x + 2}{3}$ b, $\frac{x + 3}{x}$ – $\frac{x}{x – 3}$ + ² $\frac{9}{x^2 – 3x}$

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a, $\frac{4x + 1}{2}$ – $\frac{3x + 2}{3}$
b, $\frac{x + 3}{x}$ – $\frac{x}{x – 3}$ + ² $\frac{9}{x^2 – 3x}$

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    Câu a và b

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