Решите как можно больше заданий. Очень очень нужно.

$1) \ \frac{y}{15}+3 = \frac{y+153}{45}\\ \\ \frac{y + 45}{15} - \frac{y+153}{45} = 0\\ \\ \frac{3(y+45) -y-153}{45}=0\\ \\ \frac{3y+135 -y-153}{45}=0\\ \\ \frac{2y-18}{45}=0\\ \\ 2y -18=0\\ 2y =18\\ y=9\\$

$2) \ (2\frac{1}{2}-1\frac{3}{8})*(3\frac{1}{2}-\frac{3}{6})*1\frac{1}{3}=(\frac{5}{2}-\frac{11}{8})*(\frac{7}{2}-\frac{3}{6})*\frac{4}{3}=\\ \\ =(\frac{20-11}{8})*(\frac{21-3}{6})*\frac{4}{3}=\frac{9}{8}*\frac{18}{6}*\frac{4}{3}=\frac{3}{2}*3=\frac{9}{2} = 4.5$

$3) \ (1.7:(1\frac{2}{3}-3.75)):\frac{8}{25} = 1\frac{5}{12}\\ \\ (\frac{17}{10}:(\frac{5}{3}-\frac{15}{4} )):\frac{8}{25} = \frac{17}{12}\\ \\ (\frac{17}{10}:(\frac{20x-45}{12} )):\frac{8}{25} = \frac{17}{12}\\ \\ \frac{17}{10}*(\frac{12}{5(4x-9)})*\frac{25}{8} = \frac{17}{12}\\ \\ \frac{17}{2}*(\frac{3}{4x-9} )*\frac{1}{2} = \frac{17}{12}\\ \\ \frac{51}{4(4x-9)}= \frac{17}{12}\\ \\ 51*12 = 17 * 4(4x-9)\\ 612 = 272x-612\\ 612 + 612 - 272x =0\\ 272x=1224\\ x = 1224 : 272\\ x = 4.5$