Odpowiedź :
Odpowiedź:
[tex]4 \times ( - \frac{1}{2} ) {}^{2} - 9 \times ( - \frac{1}{3} ) {}^{3} = 4 \times \frac{1}{4} - 9 \times ( - \frac{1}{27} ) = 1 + \frac{1}{3} = \frac{4}{3} [/tex]
Odpowiedź D
[tex]dla \ \x = -\frac{1}{2} \ \ i \ \ y = -\frac{1}{3}\\\\\\4x^{2}-9y^{3} = 4\cdot(-\frac{1}{2})^{2} - 9\cdot(-\frac{1}{3})^{3} = 4\cdot\frac{1}{4}-9\cdot(-\frac{1}{27}) = 1+\frac{1}{3} = \frac{3}{3}+\frac{1}{3} = \frac{4}{3}\\\\Odp. \ D.[/tex]