[tex]\displaystyle 11a).5,4+ \frac{1}{2} \\ \\5,4= \frac{54}{10} \\ \\5,4+ \frac{1}{2} = \frac{54}{10} + \frac{1}{2} = \frac{54+5}{10} = \frac{59}{10} \\ \\b).2,25+ \frac{1}{8} +1,(2) \\ \\ 2,25= \frac{225}{100} ^{(25} = \frac{9}{4} \\ \\1,(2)= \frac{12-1}{9} = \frac{11}{9} \\ \\2,25+ \frac{1}{8} +1,(2)= \frac{9}{4} + \frac{1}{8} + \frac{11}{9} = \frac{162+9+88}{72} = \frac{259}{72} [/tex]
[tex]\displaystyle c). \frac{15}{8} +0,45+ \frac{7}{4} \\ \\0,45= \frac{45}{100} ^{(5} = \frac{9}{20} \\ \\ \frac{15}{8} +0,45+ \frac{7}{4}= \frac{15}{8} + \frac{9}{20} + \frac{7}{4} = \frac{75+18+70}{40} = \frac{163}{40} \\ \\ d).0,175+ \frac{1}{16} \\ \\ 0,175= \frac{175}{1000} ^{(25} = \frac{7}{40} \\ \\ 0,175+ \frac{1}{16} = \frac{7}{40} + \frac{1}{16} = \frac{14+5}{80} = \frac{19}{80} [/tex]
[tex]\displaystyle e).0,9+ \frac{1}{6}\\ \\ 0,9= \frac{9}{10} \\ \\ 0,9+\frac{1}{6} = \frac{9}{10} + \frac{1}{6} = \frac{27+5}{30} = \frac{32}{30} ^{(2}= \frac{16}{15} \\ \\ f). \frac{7}{6} +1,45+2,55 \\ \\ 1,45= \frac{145}{100}^{(5} = \frac{29}{20} \\ \\ 2,55= \frac{255}{100} ^{(5}= \frac{51}{20} \\ \\ \frac{7}{6} +1,45+2,55= \frac{7}{6} + \frac{29}{20} + \frac{51}{20} = \frac{70+87+153}{60} = \frac{310}{60} ^{(10}= \frac{31}{6} [/tex]
[tex]\displaystyle g).1 \frac{1}{3} +0,75+ \frac{1}{9} \\ \\ 1 \frac{1}{3} = \frac{3\cdot 1+1}{3} = \frac{3+1}{3} = \frac{4}{3} \\ \\ 0,75= \frac{75}{100} ^{(25}= \frac{3}{4} \\ \\ 1 \frac{1}{3} +0,75+ \frac{1}{9} = \frac{4}{3} + \frac{3}{4} + \frac{1}{9} = \frac{48+27+4}{36} = \frac{79}{36} [/tex]
[tex]\displaystyle h).2 \frac{1}{5} +3,(4) \\ \\ 2 \frac{1}{5} = \frac{5 \cdot 2+1}{5} = \frac{10+1}{5}= \frac{11}{5} \\ \\ 3,(4)= \frac{34-3}{9} = \frac{31}{9} \\ \\ 2 \frac{1}{5} +3,(4)= \frac{11}{5} + \frac{31}{9} = \frac{99+155}{45} = \frac{254}{45} [/tex]
