Question

Математика 6 класс мерзляк номер 489​

Answer 1

$2,4;$

$\frac{15}{17};$

$\frac{7}{9};$

Пошаговое объяснение:

$1) 2+\frac{1}{1+\frac{2}{1+\frac{1}{3}}}=2+\frac{1}{1+\frac{2*3}{(1+\frac{1}{3})*3}}=2+\frac{1}{1+\frac{6}{3+1}}=2+\frac{1}{1+\frac{6}{4}}=2+\frac{1*4}{(1+\frac{6}{4})*4}=2+$

$+\frac{4}{4+6}=2+\frac{4}{10}=2+0,4=2,4;$

$2) \frac{2-\frac{\frac{1}{2}-\frac{1}{4}}{2}}{2+\frac{\frac{1}{2}-\frac{1}{4}}{2}}=\frac{2-\frac{(\frac{1}{2}-\frac{1}{4})*4}{2*4}}{2+\frac{(\frac{1}{2}-\frac{1}{4})*4}{2*4}}=\frac{2-\frac{2-1}{8}}{2+\frac{2-1}{8}}=\frac{2-\frac{1}{8}}{2+\frac{1}{8}}=\frac{(2-\frac{1}{8})*8}{(2+\frac{1}{8})*8}=\frac{16-1}{16+1}=\frac{15}{17};$

$3) \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{3}}}}=\frac{1}{2-\frac{1}{2-\frac{1*3}{(2-\frac{1}{3})*3}}}=\frac{1}{2-\frac{1}{2-\frac{3}{6-1}}}=\frac{1}{2-\frac{1}{2-\frac{3}{5}}}=\frac{1}{2-\frac{1*5}{(2-\frac{3}{5})*5}}=\frac{1}{2-\frac{5}{10-3}}=$

$=\frac{1}{2-\frac{5}{7}}=\frac{1*7}{(2-\frac{5}{7})*7}=\frac{7}{14-5}=\frac{7}{9};$