Răspuns :
Rezolvarea ecuatiei cerute, respectarea formulei pentru numetele cu perioada.
[tex](x + 0.0(5)) \div 0.(3) - 0.5 = 0.(3)[/tex]
[tex](x + \frac{5}{90} )\div \frac{3}{9} - \frac{5}{10} = \frac{3}{9} \: [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} - \frac{1}{2} = \frac{1}{3} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{1}{3} + \frac{1}{2} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{2}{6} + \frac{3}{6} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{2 + 3}{6} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{5}{6} [/tex]
[tex]x + \frac{1}{18} = \frac{5}{6} \times \frac{1}{3} [/tex]
[tex]x + \frac{1}{18} = \frac{5 \times 1}{6 \times 3} [/tex]
[tex]x + \frac{1}{18} = \frac{5}{18} [/tex]
[tex]x = \frac{5}{18} - \frac{1}{18} [/tex]
[tex]x = \frac{5 - 1}{18} [/tex]
[tex]x = \frac{4}{18} [/tex]
[tex]x = \frac{2}{9} [/tex]
[tex](x + \frac{5}{90} )\div \frac{3}{9} - \frac{5}{10} = \frac{3}{9} \: [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} - \frac{1}{2} = \frac{1}{3} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{1}{3} + \frac{1}{2} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{2}{6} + \frac{3}{6} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{2 + 3}{6} [/tex]
[tex](x + \frac{1}{18} ) \div \frac{1}{3} = \frac{5}{6} [/tex]
[tex]x + \frac{1}{18} = \frac{5}{6} \times \frac{1}{3} [/tex]
[tex]x + \frac{1}{18} = \frac{5 \times 1}{6 \times 3} [/tex]
[tex]x + \frac{1}{18} = \frac{5}{18} [/tex]
[tex]x = \frac{5}{18} - \frac{1}{18} [/tex]
[tex]x = \frac{5 - 1}{18} [/tex]
[tex]x = \frac{4}{18} [/tex]
[tex]x = \frac{2}{9} [/tex]