jika f(x) = 2x + 1 / x -3 dan g(x) 4-2x / x +2 tentukan (f°g)(x)!

[tex]f(x) = \frac{2x + 1}{x - 3} \\ g(x) = \frac{4 - 2x}{x + 2} \\ fog(x) = f(g(x)) \\ fog(x) = f( \frac{4 - 2x}{x + 2} ) \\ fog(x) = \frac{2( \frac{4 - 2x}{x + 2}) + 1 }{( \frac{4 - 2x }{x + 2}) - 3} \\ fog(x) = \frac{ \frac{2(4 - 2x)}{x + 2} + \frac{x + 2}{x + 2} }{ \frac{4 - 2x}{x + 2} - 3( \frac{x + 2}{x + 2} )} \\ fog(x) = \frac{ \frac{2(4 - 2x)}{x + 2} + \frac{x + 2}{x + 2} }{ \frac{4 - 2x}{x + 2} - \frac{3(x + 2)}{x + 2} } \\ fog(x) = \frac{ \frac{8 - 4x}{x + 2} + \frac{x + 2}{x + 2} }{ \frac{4 - 2x}{x + 2} - \frac{3x + 6}{x + 2} } \\ fog(x) = \frac{ \frac{8 - 4x + x + 2}{x + 2} }{ \frac{4 - 2x - (3x + 6)}{x + 2} } \\ fog(x) = \frac{ \frac{8 - 4x + x + 2}{x + 2} }{ \frac{4 - 2x - 3x - 6}{x + 2} } \\ fog(x) = \frac{ \frac{10 - 3x}{x + 2} }{ \frac{ - 2 - 5x}{x + 2} } \\ fog(x) = \frac{10 - 3x}{x + 2} \div \frac{ - 2 - 5x}{x + 2} \\ fog(x) = \frac{10 - 3x}{x + 2} \times \frac{x + 2}{ - 2 - 5x} \\ fog(x) = \frac{10 - 3x}{ - 2 - 5x} [/tex]
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