Odpowiedź:
[tex]5\sqrt{2}(\sqrt{2}-x)-3x(x^2-2\sqrt{2})=5\sqrt{4}-5\sqrt{2}x-3x^3+6\sqrt{2}x=5\cdot2+\sqrt{2}x-3x^3=\\\\=10+\sqrt{2}x-3x^3\\\\\\dla\ \ x=-\sqrt{2}\\\\10+\sqrt{2}x-3x^3=10+\sqrt{2}\cdot(-\sqrt{2})-3\cdot(-\sqrt{2})^3=10-\sqrt{4}-3\cdot(-\sqrt{8})=\\\\=10-2+3\sqrt{8}=8+3\sqrt{4\cdot2}=8+3\cdot2\sqrt{2}=8+6\sqrt{2}\\\\\\Odp.Warto\'s\'c\ \ wyra\.zenia\ \ jest\ \ r\'owna\ \ 6\sqrt{2}+8[/tex]
