Question

Помогите пожалуйста! Нужно превратить числа 0,(42); 1 3/7*0,2(3); 0,(63)/1,(18)*91 в дроби, используя бесконечную спадную геометрическую прогрессию. Даю 31 бал ​

Answer 1

1.

$\displaystyle\tt 0.(42)=0.42+0.0042+0.000042+...\\b_1=0.42; \ \ b_2=0.0042\\\\q=\frac{b_2}{b_1}=\frac{0.0042}{0.42}=0.01\\\\S=\frac{b_1}{1-q}=\frac{0.42}{1-0.01}=\frac{0.42}{0.99}=\frac{42}{99}=\frac{14}{33}\\\\\bold{0.(42)=\frac{14}{33}}$

Ответ: 14/33

2.

$\displaystyle\tt 0.2(3)=0.2+0.03+0.003+0.0003+...\\b_1=0.03; \ \ b_2=0.003\\\\q=\frac{b_2}{b_1}=\frac{0.003}{0.03}=0.1\\\\S=\frac{b_1}{1-q}=\frac{0.03}{1-0.1}=\frac{0.03}{0.9}=\frac{3}{90}=\frac{1}{30}\\\\\bold{0.2(3)=0.2+\frac{1}{30}=\frac{2}{10}+ \frac{1}{30}=\frac{7}{30}}\\\\\\ 1\frac{3}{7}\cdot\frac{7}{30}=\frac{10}{7}\cdot\frac{7}{30}=\frac{1}{3}$

Ответ: 1/3

3.

$\displaystyle\tt 0.(63)=0.63+0.0063+0.000063+...\\b_1=0.63; \ \ b_2=0.0063\\\\q=\frac{b_2}{b_1}=\frac{0.0063}{0.63}=0.01\\\\S=\frac{b_1}{1-q}=\frac{0.63}{1-0.01}=\frac{0.63}{0.99}=\frac{63}{99}=\frac{7}{11}\\\\\bold{0.(63)=\frac{7}{11}} \\\\\\ 1.(18)=1+0.18+0.0018+0.000018+...\\b_1=0.18; \ \ b_2=0.0018\\\\q=\frac{b_2}{b_1}=\frac{0.0018}{0.18}=0.01\\\\S=\frac{b_1}{1-q}=\frac{0.18}{1-0.01}=\frac{0.18}{0.99}=\frac{18}{99}=\frac{2}{11}\\\\\bold{1.(18)=1+\frac{2}{11}=\frac{13}{11}}$

$\displaystyle\tt\frac{0.(63)}{1.(18)}\cdot91=\frac{7}{11}:\frac{13}{11}\cdot91=\frac{7}{11}\cdot\frac{11}{13}\cdot\frac{91}{1}=7\cdot7=49$

Ответ: 49.