Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\dfrac{3\frac12\cdot(2\frac23-1\frac5{12})-\frac38}{\frac34-\frac34\cdot2\frac13}=\dfrac{\frac72\cdot(2\frac8{12}-1\frac5{12})-\frac38}{\frac34-\frac34\cdot\frac73}=\dfrac{\frac72\cdot1\frac3{12}-\frac38}{\frac34-\frac74}=\\=\dfrac{\frac72\cdot1\frac14-\frac38}{-\frac44}=\dfrac{\frac72\cdot\frac54-\frac38}{-1}=-(\dfrac{35}8-\dfrac38})=-\dfrac{32}{8}=-4\in Z[/tex]
