$1+\dfrac{1}{6x}+\dfrac{1}{x^2+3x}$
I keep getting answer like $\dfrac{6x^2 + 19x + 9}{6x(x+3)}$ but I think it's wrong because I get different answers. Please help me
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2026-03-27 05:05:57.1774587957
Write the following as a fraction
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\begin{align} 1 + \frac{1}{6x} + \frac{1}{x^2+3x} &= 1 + \frac{1}{6 x} + \frac{1}{x(x+3)} \\ &= 1 + \frac{x + 3}{6x(x+3)} + \frac{6}{6x(x+3)} = 1 + \frac{x + 9}{6x(x+3)} \\ &= \frac{6x^2 + 19x + 9}{6x(x+3)} \end{align}