Odpowiedź:
[tex]a)\ \ (3x^2+2x-4)x^2=3x^2\cdot x^2+2x\cdot x^2-4x^2=3x^4+2x^3-4x^2\\\\\\b)\ \ -2x^3(\frac{1}{4}x^2-\frac{1}{2}x-2)=-\not2^1x^3\cdot\frac{1}{\not4_{2}}x^2-\not2x^3\cdot(-\frac{1}{\not2}x)-2x^3\cdot(-2)=\\\\=-x^3\cdot\frac{1}{2}x^2+x^3\cdot x+4x^3= -\frac{1}{2}x^5+x^4+4x^3\\\\\\c)\ \ -\frac{1}{2}x^2(4x^2y-2xy^2)=-\frac{1}{\not2_{1}}x^2\cdot\not4^2x^2y-\frac{1}{\not2}x^2\cdot(-\not2xy^2)=-x^2\cdot2x^2y+x^3y^2=\\\\=-2x^4y+x^3y^2[/tex]
[tex]d)\ \ \sqrt{2}x^4(\sqrt{8}xy+y^2)=\sqrt{2}x^4\cdot\sqrt{8}xy+\sqrt{2}x^4\cdot y^2=\sqrt{16}x^5y+\sqrt{2}x^4y^2=\\\\=4x^5y+\sqrt{2}x^4y^2\\\\\\e)\ \ (x^2y-xy+xy^2)x^2y=x^2y\cdot x^2y-xy\cdot x^2y+xy^2\cdot x^2y=x^4y^2-x^3y^2+x^3y^3\\\\\\f)\ \ 2\sqrt{3}xy^3(\sqrt{3}x^2y^2-\frac{1}{2}x^3y)=2\sqrt{3}xy^3\cdot\sqrt{3}x^2y^2-\not2\sqrt{3}xy^3\cdot\frac{1}{\not2}x^3y=\\\\=2\cdot3x^3y^5-\sqrt{3}xy^3\cdot x^3y=6x^3y^5-\sqrt{3}x^4y^4[/tex]
