28. Решите уравнения

Решение:

А) \( ((3,5+1,2x):1,4-0,81) \cdot 100 = 229 \)

\( (3,5+1,2x):1,4-0,81 = \frac{229}{100} \)

\( (3,5+1,2x):1,4-0,81 = 2,29 \)

\( (3,5+1,2x):1,4 = 2,29 + 0,81 \)

\( (3,5+1,2x):1,4 = 3,1 \)

\( 3,5+1,2x = 3,1 \cdot 1,4 \)

\( 3,5+1,2x = 4,34 \)

\( 1,2x = 4,34 - 3,5 \)

\( 1,2x = 0,84 \)

\( x = \frac{0,84}{1,2} \)

\( x = 0,7 \)

Б) \( (1,2-x) \cdot 3,4 + 5,5 = 9,07 \)

\( (1,2-x) \cdot 3,4 = 9,07 - 5,5 \)

\( (1,2-x) \cdot 3,4 = 3,57 \)

\( 1,2 - x = \frac{3,57}{3,4} \)

\( 1,2 - x = 1,05 \)

\( x = 1,2 - 1,05 \)

\( x = 0,15 \)

В) \( (3\frac{1}{3}x - 2\frac{5}{7}) : 2,5 = \frac{86}{105} \)

\( (\frac{10}{3}x - \frac{19}{7}) : \frac{5}{2} = \frac{86}{105} \)

\( \frac{10}{3}x - \frac{19}{7} = \frac{86}{105} \cdot \frac{5}{2} \)

\( \frac{10}{3}x - \frac{19}{7} = \frac{86 \cdot 5}{105 \cdot 2} = \frac{430}{210} = \frac{43}{21} \)

\( \frac{10}{3}x = \frac{43}{21} + \frac{19}{7} \)

\( \frac{10}{3}x = \frac{43}{21} + \frac{19 \cdot 3}{7 \cdot 3} = \frac{43}{21} + \frac{57}{21} = \frac{100}{21} \)

\( x = \frac{100}{21} \cdot \frac{3}{10} \)

\( x = \frac{100 \cdot 3}{21 \cdot 10} = \frac{10 \cdot 1}{7 \cdot 1} = \frac{10}{7} \)

Г) \( 2(2(2(x-1)-1)-1)-1 = 57 \)

\( 2(2(2(x-1)-1)-1) = 57 + 1 \)

\( 2(2(2(x-1)-1)-1) = 58 \)

\( 2(2(x-1)-1)-1 = \frac{58}{2} \)

\( 2(2(x-1)-1)-1 = 29 \)

\( 2(2(x-1)-1) = 29 + 1 \)

\( 2(2(x-1)-1) = 30 \)

\( 2(x-1)-1 = \frac{30}{2} \)

\( 2(x-1)-1 = 15 \)

\( 2(x-1) = 15 + 1 \)

\( 2(x-1) = 16 \)

\( x-1 = \frac{16}{2} \)

\( x-1 = 8 \)

\( x = 8 + 1 \)

\( x = 9 \)

Д) \( 7\frac{5}{6} - 3\frac{6}{7}(2-3x) = 1\frac{29}{105} \)

\( \frac{47}{6} - \frac{27}{7}(2-3x) = \frac{134}{105} \)

\( -\frac{27}{7}(2-3x) = \frac{134}{105} - \frac{47}{6} \)

\( -\frac{27}{7}(2-3x) = \frac{134 \cdot 2}{105 \cdot 2} - \frac{47 \cdot 35}{6 \cdot 35} \)

\( -\frac{27}{7}(2-3x) = \frac{268}{210} - \frac{1645}{210} \)

\( -\frac{27}{7}(2-3x) = \frac{268 - 1645}{210} \)

\( -\frac{27}{7}(2-3x) = \frac{-1377}{210} \)

\( 2-3x = \frac{-1377}{210} \cdot \frac{-7}{27} \)

\( 2-3x = \frac{1377 \cdot 7}{210 \cdot 27} = \frac{1377 \cdot 1}{30 \cdot 27} = \frac{1377}{810} = \frac{153}{90} = \frac{17}{10} = 1,7 \)

\( -3x = 1,7 - 2 \)

\( -3x = -0,3 \)

\( x = \frac{-0,3}{-3} \)

\( x = 0,1 \)

Е) \( (x:\frac{2}{5} - 8\frac{11}{14}) \cdot (\frac{98}{99} + \frac{5}{99}) = 6 \)

\( (x:\frac{2}{5} - \frac{112+11}{14}) \cdot (\frac{98+5}{99}) = 6 \)

\( (x \cdot \frac{5}{2} - \frac{123}{14}) \cdot (\frac{103}{99}) = 6 \)

\( \frac{5x}{2} - \frac{123}{14} = 6 \cdot \frac{99}{103} \)

\( \frac{5x}{2} - \frac{123}{14} = \frac{594}{103} \)

\( \frac{5x}{2} = \frac{594}{103} + \frac{123}{14} \)

\( \frac{5x}{2} = \frac{594 \cdot 14 + 123 \cdot 103}{103 \cdot 14} = \frac{8316 + 12669}{1442} = \frac{20985}{1442} \)

\( x = \frac{20985}{1442} \cdot \frac{2}{5} = \frac{20985 \cdot 2}{1442 \cdot 5} = \frac{4197 \cdot 2}{1442} = \frac{4197}{721} \)

Ответ: А) 0,7; Б) 0,15; В) \(\frac{10}{7}\); Г) 9; Д) 0,1; Е) \(\frac{4197}{721}\).