Odpowiedź:
[tex]9^{2}*(-\frac{2}{3})^{3} * \frac{1}{2}-(-0,5)^{2} \\ \\9^{2}*(-(\frac{2}{3})^{3})*\frac{1}{2}-(-\frac{1}{2})^{2}\\ \\ -(3^{2})^{2}*\frac{2^{3}}{3^{3}}*\frac{1}{2}-\frac{1}{4}\\ \\ -3^{4}*\frac{2^{2}}{3^{3}}-\frac{1}{4}\\ \\ -3*2^{2}-\frac{1}{4}\\ \\ -3*4-\frac{1}{4}\\ \\ -12-\frac{1}{4}= -\frac{49}{4}[/tex]
