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maks197457 [2]
4 years ago
13

X^5y^3/x^2-x-6 * x^2 -9/x^2y^4

Mathematics
1 answer:
zvonat [6]4 years ago
6 0

\bf \cfrac{x^5y^3}{x^2-x-6}\cdot \cfrac{x^2-9}{x^2y^4}\implies \cfrac{x^5y^3}{x^2-x-6}\cdot \cfrac{\stackrel{\textit{difference of squares}}{x^2-3^2}}{x^2y^4} \\\\\\ \cfrac{x^5y^3}{~~\begin{matrix} (x-3) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x+2)}\cdot \cfrac{~~\begin{matrix} (x-3) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x+3)}{x^2y^4}\implies \cfrac{x^5y^3}{x+2}\cdot \cfrac{x+3}{x^2y^4}

\bf \cfrac{x^5y^3}{x^2y^4}\implies \cfrac{x^5x^{-2}}{y^4y^{-3}}\cdot \cfrac{x+3}{x+2}\implies \cfrac{x^{5-2}}{y^{4-3}}\cdot \cfrac{x+3}{x+2}\implies \cfrac{x^3(x+3)}{y(x+2)}\implies \cfrac{x^4+3x^3}{yx+2y}

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