Odpowiedź :
[tex]dla \ \ x = -\frac{1}{3} \ \ i \ \ y = -\frac{1}{2}\\\\\\12x^{2}-12y^{3}=12\cdot(-\frac{1}{3})^{2} - 12\cdot(-\frac{1}{2})^{3} = 12\cdot\frac{1}{9}-12\cdot(-\frac{1}{8})=\frac{12}{9}+\frac{12}{8}=\frac{4}{3}+\frac{3}{2}=\\\\=\frac{8}{6}+\frac{9}{6} = \frac{17}{6}\\\\Odp. \ C.[/tex]
Odpowiedź:
to będzie
[tex] \frac{17}{6} [/tex]
bo
[tex]12 \times ( - \frac{1}{3} ) ^{2} - 12 \times {( - \frac{1}{2} })^{3} = 12 \times \frac{1}{9} - 12 \times ( - \frac{1}{8} ) = \frac{12}{9} + \frac{12}{8} = \frac{4}{3} + \frac{3}{2} = \frac{8}{6} + \frac{9}{6} = \frac{17}{6} [/tex]