Question

simplifique as frações algébricas:​

Answer 1

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$\tt{a)}~\sf{\dfrac{\diagdown\!\!\!\!\!\!42^7\diagup\!\!\!n^3n^3}{\diagdown\!\!\!\!\!\!18_3m\diagup\!\!\!n^2}=\dfrac{7n^4}{3m}}\\\tt{b)}~\sf{\dfrac{8x-16y}{4x-8y}=\dfrac{\backslash\!\!\!8^2\cdot\diagup\!\!\!(x\diagup\!\!\!-\diagup\!\!\!\2y)}{\backslash\!\!\!4\cdot\diagup\!\!\!(x\diagup\!\!\!-\diagup\!\!\!2y)}=2}$

$\tt{c)}~\sf{\dfrac{x^3-x^2-4x+4}{x^2-4}=\dfrac{x^2\cdot(x-1)-4\cdot(x-1)}{x^2-4}}\\\sf{\dfrac{\diagup\!\!\!\!(x^2\diagup\!\!\!-\diagup\!\!\!4)\cdot(x-1)}{\diagup\!\!\!(x^2\diagup\!\!\!-\diagup\!\!\!4)}=x-1}$

$\tt{d)}~\sf{\dfrac{x^2-64}{x^2-16x+64}=\dfrac{\diagup\!\!\!(x\diagup\!\!\!-\diagup\!\!\!8)\cdot\diagup\!\!\!(x+8)}{\diagup\!\!\!(x\diagup\!\!\!-\diagup\!\!\!8)^2}=\dfrac{x+8}{x-8}}$