1.315. Решите уравнение: А) ((3,5 + 1,2x) : 1,4 - 0,81) * 100 = 229

Привет! Давай решим уравнения из задания 1.315. А) $((3,5 + 1,2x) : 1,4 - 0,81) \cdot 100 = 229$ 1) $(3,5 + 1,2x) : 1,4 - 0,81 = 2,29$ 2) $(3,5 + 1,2x) : 1,4 = 3,1$ 3) $3,5 + 1,2x = 4,34$ 4) $1,2x = 0,84$ 5) $x = 0,7$ Б) $(1,2 - x) \cdot 3,4 + 5,5 = 9,07$ 1) $(1,2 - x) \cdot 3,4 = 3,57$ 2) $1,2 - x = 1,05$ 3) $x = 0,15$ В) $(3\frac{1}{3}x - 2\frac{5}{7}) : 2,5 = \frac{86}{105}$ Переведем в неправильные дроби: $(\frac{10}{3}x - \frac{19}{7}) : \frac{5}{2} = \frac{86}{105}$ 1) $\frac{10}{3}x - \frac{19}{7} = \frac{86}{105} \cdot \frac{5}{2} = \frac{43}{21}$ 2) $\frac{10}{3}x = \frac{43}{21} + \frac{19}{7} = \frac{43+57}{21} = \frac{100}{21}$ 3) $x = \frac{100}{21} \cdot \frac{3}{10} = \frac{10}{7} = 1\frac{3}{7}$ Г) $2(2(2(x-1)-1)-1)=57$ 1) $2(2(x-1)-1)-1 = 28,5$ 2) $2(2(x-1)-1) = 29,5$ 3) $2(x-1)-1 = 14,75$ 4) $2(x-1) = 15,75$ 5) $x-1 = 7,875$ 6) $x = 8,875$ Д) $7\frac{5}{6} - 3\frac{6}{7}(2-3x) = 1\frac{29}{105}$ $\frac{47}{6} - \frac{27}{7}(2-3x) = \frac{134}{105}$ 1) $\frac{27}{7}(2-3x) = \frac{47}{6} - \frac{134}{105} = \frac{1645 - 268}{210} = \frac{1377}{210} = \frac{459}{70}$ 2) $2-3x = \frac{459}{70} : \frac{27}{7} = \frac{459}{70} \cdot \frac{7}{27} = \frac{17}{10} = 1,7$ 3) $3x = 0,3$ 4) $x = 0,1$ Е) $(x : 1\frac{2}{5} - 8\frac{11}{14}) \cdot \frac{98}{99} + \frac{5}{99} = 6$ $(x : 1,4 - \frac{123}{14}) \cdot \frac{98}{99} = \frac{589}{99}$ 1) $x : 1,4 - \frac{123}{14} = \frac{589}{99} : \frac{98}{99} = \frac{589}{98}$ 2) $x : 1,4 = \frac{589}{98} + \frac{861}{98} = \frac{1450}{98} = \frac{725}{49}$ 3) $x = \frac{725}{49} \cdot 1,4 = \frac{725}{49} \cdot \frac{7}{5} = \frac{145}{7} = 20\frac{5}{7}$