Respuesta :

[tex]\bf \cfrac{-14x^3}{x^3-5x^4}\implies \cfrac{-14\underline{x^3}}{\underline{x^3}(1-5x)}\implies \cfrac{-14}{1-5x}\implies \cfrac{14}{5x-1}\\\\ -------------------------------\\\\ \cfrac{x}{4x+x^2}\implies \cfrac{\underline{x}}{\underline{x}(4+x)}\implies \cfrac{1}{4+x}[/tex]
tramserran
(-14x³)/(x³ - 5x^4)
factor out x³ from the numerator (top) and also from the denominator (bottom):
-14 (x³) / (1-5x)(x³)
cancel out the common factor of x³:
-14/(1-5x)

Do the same process for the second question (but factor out x):
x/(4x + x²) = 1/(4+x)

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