Помогите решить хотя бы 1 номер

НОМЕР 1
$\frac{17}{31}$ *17 = $\frac{17*17}{31*1}$ = $\frac{289}{31}$ = 9 $\frac{10}{31}$
5: $\frac{3}{5}$ = $\frac{5*5}{3}$ = $\frac{25}{3}$ = 8 $\frac{1}{3}$
$\frac{6}{7} : 12 = \frac{6*1}{7*12} = \frac{1}{14}$
$\frac{4}{9}: \frac{27}{36} = \frac{4*36}{9*27} = \frac{16}{27}$

НОМЕР 2
$\frac{2}{7} = \frac{7}{2} = 3,5$
$\frac{1}{2} = 3,5 \frac{1}{9} = \frac{9}{1} = 9$
$25 = \frac{1}{25} = 0,04$
$\frac{1}{n} = \frac{n}{1} = n$

НОМЕР 3
$\frac{13}{18}* \frac{7}{17} + \frac{2}{9} * \frac{7}{17} = \frac{7}{17} * ( \frac{13}{18} + \frac{2}{9} ) = \frac{7}{17} * ( \frac{13}{18} + \frac{4}{18} ) = \frac{7}{17} * \frac{17}{18} ) = \frac{7*17}{17*18}$ = $\frac{7}{18}$
$( \frac{5}{11} - \frac{3}{22} )*44 = ( \frac{10}{22} - \frac{3}{22} )*44 = \frac{7}{22}*44 = \frac{7*44}{22} = 14$
$\frac{19}{20}:\frac{38}{5}*\frac{5}{16} = \frac{19*5}{20*38}*\frac{5}{16} = \frac{1}{8}* \frac{5}{16} = \frac{1*5}{8*16} = \frac{5}{128}$

НОМЕР 4
$\frac{17}{29}*4 = \frac{17*4}{29} = \frac{68}{29} = 2 \frac{10}{29}$ (м) - 2-ая сторона треугольника
$\frac{143}{29} - ( \frac{17}{29} + 2 \frac{10}{29} ) = \frac{143}{29} - 2 \frac{27}{29} = \frac{143}{29} - \frac{85}{29} = \frac{58}{29} = 2$ (м)
Ответ: 3-я сторона треугольника равна 2 м.