Question

Operaciones entre expresiones algebraicas​

Answer 1

Respuesta:

Resolvemos:

$P_{1} =\frac{1}{3}x^3- \frac{35}{36} x^2y+\frac{2}{3}xy^2 -\frac{3}{8} y^3$

$P_{2} =\frac{1}{2}x^2- \frac{1}{3}xy +\frac{1}{4}y^2$

Operamos:

$P_{1}+ P_{2}$

$\frac{1}{3}x^3-\:\frac{35}{36}\:x^2y+\frac{2}{3}xy^2\:-\frac{3}{8}\:y^3+\frac{1}{2}x^2-\:\frac{1}{3}xy\:+\frac{1}{4}y^2$

$\frac{x^3+2xy^2-xy}{3}-\frac{35x^2y}{36}-\frac{3y^3}{8}+\frac{x^2}{2}+\frac{y^2}{4}$

$P_{1}- P_{2}$

$\frac{1}{3}x^3-\:\frac{35}{36}\:x^2y+\frac{2}{3}xy^2\:-\frac{3}{8}\:y^3-\left(\frac{1}{2}x^2-\:\frac{1}{3}xy\:+\frac{1}{4}y^2\right)$

$\frac{x^3}{3}-\frac{35x^2y}{36}+\frac{2xy^2}{3}-\frac{3y^3}{8}-\left(\frac{x^2}{2}+\frac{y^2}{4}-\frac{xy}{3}\right)$

$\frac{x^3+2xy^2}{3}-\frac{35x^2y}{36}-\frac{3y^3}{8}-\frac{x^2}{2}+\frac{xy}{3}-\frac{y^2}{4}$

$\frac{x^3+2xy^2+xy}{3}-\frac{35x^2y}{36}-\frac{3y^3}{8}-\frac{x^2}{2}-\frac{y^2}{4}$

$P_{1}*P_{2}$

$\left(\frac{1}{3}x^3-\:\frac{35}{36}\:x^2y+\frac{2}{3}xy^2\:-\frac{3}{8}\:y^3\right)\left(\frac{1}{2}x^2-\:\frac{1}{3}xy\:+\frac{1}{4}y^2\right)$

$\left(\frac{2xy^2}{3}+\frac{x^3}{3}-\frac{35x^2y}{36}-\frac{3y^3}{8}\right)\left(\frac{1}{2}x^2-\frac{1}{3}xy+\frac{1}{4}y^2\right)$

$\left(-\frac{3y^3}{8}+\frac{x^3+2xy^2}{3}-\frac{35x^2y}{36}\right)\left(\frac{x^2}{2}+\frac{y^2}{4}-\frac{xy}{3}\right)$

$P_{1}/ P_{2}$

$\frac{\left(\frac{1}{3}x^3-\:\frac{35}{36}\:x^2y+\frac{2}{3}xy^2\:-\frac{3}{8}\:y^3\right)}{\left(\frac{1}{2}x^2-\:\frac{1}{3}xy\:+\frac{1}{4}y^2\right)}$

$\frac{\frac{1}{3}x^3-\frac{35}{36}x^2y+\frac{2}{3}xy^2-\frac{3}{8}y^3}{\frac{x^2}{2}-\frac{xy}{3}+\frac{y^2}{4}}$

$\frac{\frac{x^3}{3}-\frac{35x^2y}{36}+\frac{2xy^2}{3}-\frac{3y^3}{8}}{\frac{x^2}{2}-\frac{xy}{3}+\frac{y^2}{4}}$

$\frac{24\left(x^3+2xy^2\right)-70x^2y-27y^3}{6\left(6x^2-4xy+3y^2\right)}$