132. Теңдеуді шешіңдер: 1) \(x-1=\frac{1}{1-\frac{3}{5}}\); 2) \(x-2=\frac{2}{4-1\frac{1}{2}}\); 3) \(x-3=\frac{3}{2+1\frac{1}{3}}\).

1. \(1-\frac{3}{5}=\frac{2}{5}\).

   Сонда \(x-1=\frac{1}{\frac{2}{5}}=\frac{5}{2}\), демек \(x=1+\frac{5}{2}=\frac{7}{2}\).

2. \(4-1\frac{1}{2}=\frac{5}{2}\).

   Сонда \(x-2=\frac{2}{\frac{5}{2}}=\frac{4}{5}\), демек \(x=2+\frac{4}{5}=\frac{14}{5}\).

3. \(2+1\frac{1}{3}=\frac{10}{3}\).

   Сонда \(x-3=\frac{3}{\frac{10}{3}}=\frac{9}{10}\), демек \(x=3+\frac{9}{10}=\frac{39}{10}\).

Жауабы: 1) \(x=\frac{7}{2}\); 2) \(x=\frac{14}{5}\); 3) \(x=\frac{39}{10}\).