Mn giúp em bài này vs ạ

a)
\[
\frac{x+1}{3} + \frac{2x-1}{3} = \frac{(x+1) + (2x-1)}{3} = \frac{3x}{3} = x
\]

b)
\[
\frac{5x - 2y}{x^2} + \frac{y - 4x}{x^2} = \frac{(5x - 2y) + (y - 4x)}{x^2} = \frac{x - y}{x^2}
\]

c)
\[
\frac{x - 1}{12x} + \frac{2}{12x} + \frac{7}{12x} = \frac{(x - 1 + 2 + 7)}{12x} = \frac{x + 8}{12x}
\]
\[
\frac{6 - 3x}{12x} = \frac{(6 - 3x)}{12x}
\]

d)
\[
\frac{3}{x - 4} + \frac{2}{4 - x} = \frac{3}{x - 4} - \frac{2}{x - 4} = \frac{3 - 2}{x - 4} = \frac{1}{x - 4}
\]

e)
\[
\frac{x + 2}{1 - x} + \frac{x - 9}{1 - x} = \frac{(x + 2) + (x - 9)}{1 - x} = \frac{2x - 7}{1 - x}
\]

f)
\[
\frac{x + 3}{x^2 - 1} \cdot \frac{x^2 + x}{x^2 + x} = \frac{(x + 3)(x^2 + x)}{(x - 1)(x + 1)(x^2 + x)} = \frac{x + 3}{(x - 1)(x + 1)}
\]