Respuesta :

[tex]\bf \cfrac{p^2-9p}{p^2-4p-45}-\cfrac{2}{p+5}\qquad \begin{array}{lcclll} p^2&-4p&-45\\ &\uparrow &\uparrow \\ &5-9&5\cdot -9 \end{array}\qquad thus \\\\\\ \cfrac{p\underline{(p-9)}}{\underline{(p-9)}(p+5)}-\cfrac{2}{p+5}\implies \cfrac{p}{p+5}-\cfrac{2}{p+5}\impliedby \textit{LCD of p+5} \\\\\\ \cfrac{p-2}{p+5}[/tex]
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[tex]\displaystyle\\ \frac{p^2-9p}{p^2-4p - 45} - \frac{2}{p+5}=\\\\ =\frac{p(p-9)}{p^2 \underbrace{+5p-9p}_{=-4p} - 45} - \frac{2}{p+5}=\\\\\\ =\frac{p(p-9)}{p(p+5)-9(p +5)} - \frac{2}{p+5}=\\\\ =\frac{p(p-9)}{(p+5)(p-9)} - \frac{2}{p+5}=\\\\ =\frac{p}{p+5} - \frac{2}{p+5}=\boxed{\bold{\frac{p-2}{p+5}}} [/tex]



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