Question

porfaaa ayudenmeeee les dare puntoss

Answer 1

A)

= $\frac{-8x^4y^5+12x^3y^7-10x^2y^9}{-2x^2y^3}$

= $\frac{-2x^2y^5\left(4x^2-6xy^2+5y^4\right)}{-2x^2y^3}$

= $\frac{x^2y^5\left(4x^2-6xy^2+5y^4\right)}{x^2y^3}$

= $\frac{y^5\left(4x^2-6xy^2+5y^4\right)}{y^3}$

= $\frac{y^3y^2\left(4x^2-6xy^2+5y^4\right)}{y^3}$

= $y^2\left(4x^2-6xy^2+5y^4\right)$

B)

= $\frac{-40x^9y^{11}-72x^5y^{10}-88x^2y^4}{-8x^2y^3}$

= $\frac{-\left(40x^9y^{11}+72x^5y^{10}+88x^2y^4\right)}{-8x^2y^3}$

= $\frac{40x^9y^{11}+72x^5y^{10}+88x^2y^4}{8x^2y^3}$

= $\frac{8x^2y^4\left(5x^7y^7+9x^3y^6+11\right)}{8x^2y^3}$

= $\frac{y^4\left(5x^7y^7+9x^3y^6+11\right)}{y^3}$

= $\frac{y^3y\left(5x^7y^7+9x^3y^6+11\right)}{y^3}$

= $y\left(5x^7y^7+9x^3y^6+11\right)$

C) (Esta no la se)

D)

= $\frac{\frac{7}{4}x^4y^7-\frac{9}{8}x^3y^8-\frac{11}{6}x^5y^{10}}{\frac{1}{2}xy^4}$

= $\frac{\frac{7}{4}x^4y^7-\frac{9}{8}x^3y^8-\frac{11}{6}x^5y^{10}}{\frac{xy^4}{2}}$

= $\frac{\frac{7x^4y^7}{4}-\frac{9}{8}x^3y^8-\frac{11}{6}x^5y^{10}}{\frac{xy^4}{2}}$

= $\frac{\frac{7x^4y^7}{4}-\frac{9x^3y^8}{8}-\frac{11}{6}x^5y^{10}}{\frac{xy^4}{2}}$

= $\frac{\frac{7x^4y^7}{4}-\frac{9x^3y^8}{8}-\frac{11x^5y^{10}}{6}}{\frac{xy^4}{2}}$

= $\frac{\left(\frac{7x^4y^7}{4}-\frac{9x^3y^8}{8}-\frac{11x^5y^{10}}{6}\right)\cdot \:2}{xy^4}$

= $\frac{2\cdot \frac{-44x^5y^{10}+42x^4y^7-27x^3y^8}{24}}{xy^4}$

= $\frac{\frac{42x^4y^7-27x^3y^8-44x^5y^{10}}{12}}{xy^4}$

= $\frac{42x^4y^7-27x^3y^8-44x^5y^{10}}{12xy^4}$

= $\frac{y^7x^3\left(42x-27y-44x^2y^3\right)}{12xy^4}$

= $\frac{x^2y^3\left(-44x^2y^3+42x-27y\right)}{12}$