DBH3-Zatikiak-Buruzko kalkulua

 $-\frac{3}{5}:\frac{9}{5}=$  

 $-\frac{2}{3}+\frac{1}{2}=$  

 $\frac{2}{3}-\left(\frac{-1}{9}\right)=$  

 $\frac{49}{25}\cdot\frac{5}{7}=$  

 $\frac{12}{10}\cdot\frac{5}{3}=$  

 $\frac{200}{11}\cdot\frac{22}{100}=$  

 $\left(-\frac{5}{9}\right):\left(-\frac{7}{9}\right)=$  

 $2\cdot\left(-\frac{15}{4}\right)=$  

 $2+\frac{3}{4}=$  

 $-3-\frac{2}{5}=$