Problema 69: Aritmética

Problema:

Resolver aplicando generatriz:
$$a)0,2+1,2+0,\hat{6}\\b)2,\hat{12}-1,\hat{93}$$

Solución:


$$a)0,2+1,2+0,\hat{6}\\0,2= \frac{2}{10} = \frac{1}{5}  \\ 1,2= \frac{12}{10}= \frac{6}{5}\\x=0,\hat{6}\\10x=6,\hat{6}\\10x-x=9x=6 \\ x= \frac{6}{9}= \frac{2}{3}=0,\hat{6}\\   \\ 0,2+1,2+0,\hat{6}= \frac{1}{5}+  \frac{6}{5} + \frac{2}{3}\\ =\frac{7}{5}+   \frac{2}{3}= \frac{7(3)+2(5)}{5(3)}= \frac{21+10}{15}= \frac{31}{15}  $$

$$b)2,\hat{12}-1,\hat{93}\\x=2,\hat{12}\\100x=212,\hat{12}\\100x-x=99x=212,\hat{12}-2,\hat{12}=210 \\ x= \frac{210}{99} = \frac{70}{33}  \\ y=1,\hat{93} \\ 100y=193,\hat{93} \\ 100y-y=99y=192 \\ y= \frac{192}{99}= \frac{64}{33} \\   2,\hat{12}-1,\hat{93}= \frac{70}{33}- \frac{64}{33}= \frac{6}{33}= \frac{2}{11}    $$

Saludos! y Donen si pueden