Odpowiedź:
[tex]a)\ \ 3(x^2-2x+3)-\frac{1}{2}(6x^2-4x+10)=3x^2-6x+9-3x^2+2x-5=-4x+4\\\\\\dla\ \ x=-1,5\\\\-4x+4=-4\cdot(-1,5)+4=6+4=10\\\\\\b)\ \ x(3xy+2y)-y(4x^2+2x-y)=3x^2y+2xy-4x^2y-2xy+y^2=-x^2y+y^2\\\\\\dla\ \ x=-0,2\ \ ,\ \ y=10\\\\-x^2y+y^2=-(-0,2)^2\cdot10+10^2=-0,04\cdot10+100=-0,4+100=99,6[/tex]
