Răspuns:
[tex]0.(3) = \frac{3}{9} - < > 1.(3) = 1 \frac{3}{9} = \frac{ 9 + 3}{9} = \frac{12}{3} < > 1 \frac{7}{9} = {}^{} \frac{9 + 7}{9} = \frac{16}{9} \\ ( \frac{3}{9} + \frac{12}{9} \div \frac{16}{9}) \div (1 + \frac{1}{12} ) = ( \frac{3}{9} + \frac{12}{9 {}^{(9} } \times \frac{9 {}^{(9} }{16} ) \div ( \frac{12 + 1}{12} ) \\ = ( \frac{3}{9} + 12 {}^{(4} \times \frac{1}{16 {}^{(4} }) \div ( \frac{13}{12} ) = ( \frac{3}{9} + 3 \times \frac{1}{4} ) \div ( \frac{13}{12} ) \\ = ( \frac{3 {}^{(3} }{9 {}^{(3} } + \frac{3 {}^{} }{4} ) \div ( \frac{13}{12} ) = ( \frac{1}{3} + \frac{3}{4} ) \div ( \frac{13}{12} ) \\ = ( \frac{ {}^{4)} 1}{ \: \: 3} + \frac{ {}^{3)} 3}{ \: \: 4} ) \div ( \frac{13}{12} ) = ( \frac{4 + 9}{12}) \div ( \frac{13}{12}) = ( \frac{13}{12}) \div ( \frac{13}{12}) = 1 \\ numitorul \: comun \: e \: 12[/tex]
