Odpowiedź:
[tex]a)\ \ 2x(3x-4)-6x(x^2+2x-3)=6x^2-8x-6x^3-12x^2+18x=-6x^3-6x^2+10x\\\\\\b)\ \ -\frac{1}{2}x^2(x^2-2x+6)-2x(\frac{1}{2}x^2-4x)=-\frac{1}{2}x^4+x^3-3x^2-x^3+8x^2=-\frac{1}{2}x^4+5x^2\\\\\\c)\ \ \sqrt{6}x^3(\sqrt{3}x-\sqrt{2})+\sqrt{3}x(2x^2-4\sqrt{2}x)=\sqrt{18}x^4-\sqrt{12}x^3+2\sqrt{3}x^3-4\sqrt{6}x^2=\\\\=\sqrt{9\cdot2}x^4-\sqrt{4\cdot3}x^3+2\sqrt{3}x^3-4\sqrt{6}x^2=3\sqrt{2}x^4-2\sqrt{3}x^3+2\sqrt{3}x^3-4\sqrt{6}x^2=\\\\=3\sqrt{2}x^4-4\sqrt{6}x^2[/tex]
[tex]d)\ \ xy^2(2x-3xy+y)+\frac{1}{4}x^2y(\frac{1}{2}y^2-8y)=2x^2y^2-3x^2y^3+xy^3+\frac{1}{8}x^2y^3-2x^2y^2=\\\\=-3x^2y^3+\frac{1}{8}x^2y^3+xy^3=-\frac{24}{8}x^2y^3+\frac{1}{8}x^2y^3+xy^3=-\frac{23}{8}x^2y^3+xy^3\\\\\\e)\ \ -\frac{1}{2}x^3y(xy-2xy^2)-\frac{1}{8}xy^2(4x^3-16x^2y)=-\frac{1}{2}x^4y^2+x^4y^3-\frac{1}{2}x^4y^2+2x^3y^3=\\\\=-x^4y^2+x^4y^3+2x^3y^3[/tex]
[tex]f)\ \ -\frac{1}{4}x^2y(-2x+y)-\frac{3}{4}y(x^2-x^2y)-\frac{1}{2}(x^3y+xy)=\\\\=\frac{1}{2}x^3y-\frac{1}{4}x^2y^2-\frac{3}{4}x^2y+\frac{3}{4}x^2y^2-\frac{1}{2}x^3y-\frac{1}{2}xy=\frac{2}{4}x^2y^2-\frac{3}{4}x^2y-\frac{1}{2}xy=\\\\=\frac{1}{2}x^2y^2-\frac{3}{4}x^2y-\frac{1}{2}xy[/tex]
