$[(x^2-1/2xy)(2x^2+xy)-1/2(2x^2)^2]^2:(-3/4xy^2)+x^3(1/3y-x)(3x+y)$

A cura di: Francesco Speciale

Svolgimento:
$[(x^2-1/2xy)(2x^2+xy)-1/2(2x^2)^2]^2:(-3/4xy^2)+x^3(1/3y-x)(3x+y)=$
$=[2x^4+x^3y-x^3y-1/2x^2y^2-1/2(4x^4)]^2:(-3/4xy^2)+x^3(xy+1/3y^2-3x^2-xy)=$
$=[2x^4-1/2x^2y^2-2x^4]^2:(-3/4xy^2)+1/3y^2x^3-3x^5=$
$=[-1/2x^2y^2]^2:(-3/4xy^2)+1/3y^2x^3-3x^5=$
$=[1/4x^4y^4]:(-3/4xy^2)+1/3y^2x^3-3x^5=-1/3y^2x^3+1/3y^2x^3-3x^5=-3x^5$.