Упр.534 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)

Задача

а) 1/(a+1/(a+1/(a+1))), если a = 2;
Если a=2, то получим
1/(a+1/(a+1/(a+1)))=1/(2+1/(2+1/(2+1)))=1/(2+1/(2+1/3))=1/(2+1/(2 1/3))=1/(2+1/((2•3+1)/3))=1/(2+1/(7/3))=1/(2+1:7/3)=1/(2+1•3/7)=
=1/(2+3/7)=1/(2 3/7)=1:2 3/7=1:(2•7+3)/7=1:17/7=1•7/17=7/17
б) 1/(b-1/(b-1/(b-1))), если b = 3.
Если b=3, то получим
1/(b-1/(b-1/(b-1)))=1/(3-1/(3-1/(3-1)))=1/(3-1/(3-1/2))=1/(3-1/(2 2/2-1/2))=1/(3-1/(2+(2-1)/2))=1/(3-1/(2+1/2))=1/(3-1/(2 1/2))=
=1/(3-1/((2•2+1)/2))=1/(3-1/(5/2))=1/(3-1:5/2)=1/(3-1•2/5)=1/(3-2/5)=1/(2 5/5-2/5)=1/(2+5/5-2/5)=1/(2+(5-2)/5)=
=1/(2+3/5)=1/(2 3/5)=1:2 3/5=1:(2•5+3)/5=1:13/5=1•5/13=5/13

Подробный ответ

а) Подставим $$a=2$$:

$$ \frac{1}{a+\frac{1}{a+\frac{1}{a+1}}} = \frac{1}{2+\frac{1}{2+\frac{1}{2+1}}} = \frac{1}{2+\frac{1}{2+\frac{1}{3}}} = \frac{1}{2+\frac{1}{\frac{7}{3}}} = \frac{1}{2+\frac{3}{7}} = \frac{1}{2\frac{3}{7}} = \frac{1}{\frac{17}{7}} = \frac{7}{17}. $$

б) Подставим $$b=3$$:

$$ \frac{1}{b-\frac{1}{b-\frac{1}{b-1}}} = \frac{1}{3-\frac{1}{3-\frac{1}{3-1}}} = \frac{1}{3-\frac{1}{3-\frac{1}{2}}} = \frac{1}{3-\frac{1}{\frac{5}{2}}} = \frac{1}{3-\frac{2}{5}} = \frac{1}{2\frac{3}{5}} = \frac{1}{\frac{13}{5}} = \frac{5}{13}. $$