Question

-[-(x²-y²)+2x²-3y²-(x²-2x²-3y2)]​

Answer 1

Respuesta:

$\boxed{\mathrm{-\left[-\left(x^2-y^2\right)+2x^2-3y^2-\left(x^2-2x^2-3y^2\right)\right]}}$

Resolvemos

$\mathrm{-\left[-\left(x^2-y^2\right)+2x^2-3y^2-\left(x^2-2x^2-3y^2\right)\right]}$

$\mathrm{-\left(x^2-y^2\right)+2x^2-3y^2-\left(x^2-2x^2-3y^2\right)=2x^2+y^2}$

$\mathrm{-\left(2x^2+y^2\right)=-\left(2x^2\right)-\left(y^2\right)=-2x^2-y^2}$

Solución:  - 2x² - y²

Saludos...