Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 2

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\[1)\ \frac{2^{\backslash 9}}{11} + \frac{1^{\backslash 11}}{9} = \frac{2 \bullet 9 + 11}{99} =\]

\[= \frac{18 + 11}{99} = \frac{29}{99} = 0,(29)\]

\[2)\ \frac{8^{\backslash 3}}{13} + \frac{2^{\backslash 13}}{3} = \frac{8 \bullet 3 + 2 \bullet 13}{39} =\]

\[= \frac{24 + 26}{39} = \frac{50}{39} = 1\frac{11}{39} =\]

\[= 1,(282051)\]

\[3)\ \frac{1}{3} + 1,25 = \frac{1}{3} + 1\frac{25}{100} =\]

\[= \frac{1^{\backslash 4}}{3} + \frac{4 + 1^{\backslash 3}}{4} = \frac{4 + 5 \bullet 3}{12} =\]

\[= \frac{19}{12} = 1\frac{7}{12} = 1,58(3)\]

\[4)\ \frac{1}{6} + 0,33 = \frac{1^{\backslash 100}}{6} + \frac{33^{\backslash 6}}{100} =\]

\[= \frac{100 + 33 \bullet 6}{600} = \frac{100 + 198}{600} =\]

\[= \frac{298}{600} = \frac{149}{300} = 0,49(6)\]

\[5)\ \frac{3}{14} \bullet 1,05 = \frac{3}{14} \bullet 1\frac{5}{100} =\]

\[= \frac{3}{14} \bullet \frac{20 + 1}{20} = \frac{3}{14} \bullet \frac{21}{20} =\]

\[= \frac{3}{2} \bullet \frac{3}{20} = \frac{9}{40} = 0,225\]

\[6)\ \frac{7}{9} \bullet 1,7 = \frac{7}{9} \bullet 1\frac{7}{10} =\]

\[= \frac{7}{9} \bullet \frac{10 + 7}{10} = \frac{7}{9} \bullet \frac{17}{10} = \frac{119}{90} =\]

\[= 1\frac{29}{90} = 1,3(2)\]