Question

a)[1,3(5)-0,0(2)+0,6]-(1 intreg 7supra 15 -1 supra 5)

Answer 1

[tex]1,3(5)= \frac{61}{45} \\ 0,0(2)= \frac{1}{45} \\ 0,6= \frac{3}{5} \\ 1 \frac{7}{15}= \frac{22}{15} \\ \\( \frac{61}{45} -\frac{1}{45} +\frac{3}{5})-(\frac{22}{15} -\frac{1}{5})= \\ \\ (\frac{61-1+27}{45})-(\frac{22-3}{15})= \\ \\ \frac{87}{45}-\frac{19}{15}= \frac{87-57}{45}=\frac{30}{45}= \frac{2}{3}. [/tex]

/robèrtinni