[tex]\displaystyle\bf\\a)\\\\\frac{5}{11}~din~\left(\frac{59}{3}+\frac{40}{3}\right)=\\\\\\=\frac{5}{11}\times\frac{59+40}{3}=\\\\\\=\frac{5}{11}\times\frac{99}{3}=\\\\\\=\frac{5\times99}{11\times3}=\frac{5\times9}{3}=5\times3=\boxed{\bf15}\\\\\\b)\\\\\frac{7}{9}~din~\left(\frac{117}{4}-\frac{9}{4}\right)=\\\\\\=\frac{7}{9}\times\frac{117-9}{4}=\\\\\\=\frac{7}{9}\times\frac{108}{4}=\\\\\\=\frac{7\times108}{9\times4}=\frac{7\times12}{4}=7\times3=\boxed{\bf21}[/tex]
