Respuesta :

Julik
[tex] \frac{3xy^2z + 2x^2y^3}{3x^2yz} = 100 \\ \frac{xy^2(3z + 2xy)}{3x^2yz} = 100 \\ \frac{y(3z + 2xy)}{3xz} = 100 \\ y(3z + 2xy)=100 \times 3xz \\ y(3z + 2xy)=300xz \\ 3zy + 2xy^2-300xz=0 \\ 2xy^2+3zy-300xz=0 \\ \Delta=(3z)^2-4 \times 2x \times(-300xz) =9z^2+2400x^2z \\ y= \frac{-3z \pm \sqrt{9z^2+2400x^2z } }{2 \times 2x} = \frac{-3z \pm \sqrt{3z(3z+800x^2) } }{4x} [/tex]