28. Решите уравнения

Решение:

A) ((3,5+1,2x):1,4-0,81)·100 = 229

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

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

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

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

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

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

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

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

\( x = \frac{0,84}{1,2} = \frac{8,4}{12} = 0,7 \)

Б) (1,2-х)·3,4+5,5 = 9,07

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

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

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

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

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

\( x = 0,15 \)

B) (3¹/₃x-2⁵/₇):2,5 = ⁸⁶/₁₀₅

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

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

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

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

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

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

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

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) 7⁵/₆ - 3⁵/₇(2-3x) = 1²⁹/₁₀₅

\( \frac{47}{6} - \frac{26}{7}(2-3x) = \frac{134}{105} \)

\( \frac{26}{7}(2-3x) = \frac{47}{6} - \frac{134}{105} \)

\( \frac{26}{7}(2-3x) = \frac{47 \cdot 35}{6 \cdot 35} - \frac{134 \cdot 2}{105 \cdot 2} = \frac{1645}{210} - \frac{268}{210} \)

\( \frac{26}{7}(2-3x) = \frac{1645 - 268}{210} = \frac{1377}{210} \)

\( 2-3x = \frac{1377}{210} : \frac{26}{7} = \frac{1377}{210} \cdot \frac{7}{26} \)

\( 2-3x = \frac{1377 \cdot 7}{210 \cdot 26} = \frac{1377 \cdot 1}{30 \cdot 26} = \frac{1377}{780} \)

\( 2-3x = \frac{459}{260} \)

\( 3x = 2 - \frac{459}{260} = \frac{2 \cdot 260}{260} - \frac{459}{260} = \frac{520 - 459}{260} = \frac{61}{260} \)

\( x = \frac{61}{260} : 3 = \frac{61}{260 \cdot 3} = \frac{61}{780} \)

E) (x:1²/₅ - 8¹¹/₁₄)·⁹⁸/₉₉ + ⁵/₆ = 6

\( (x : \frac{7}{5} - \frac{123}{14}) \cdot \frac{98}{99} = 6 - \frac{5}{6} \)

\( (\frac{5x}{7} - \frac{123}{14}) \cdot \frac{98}{99} = \frac{36 - 5}{6} = \frac{31}{6} \)

\( \frac{5x}{7} - \frac{123}{14} = \frac{31}{6} : \frac{98}{99} = \frac{31}{6} \cdot \frac{99}{98} \)

\( \frac{5x}{7} - \frac{123}{14} = \frac{31 \cdot 99}{6 \cdot 98} = \frac{31 \cdot 33}{2 \cdot 98} = \frac{1023}{196} \)

\( \frac{5x}{7} = \frac{1023}{196} + \frac{123}{14} \)

\( \frac{5x}{7} = \frac{1023}{196} + \frac{123 \cdot 14}{14 \cdot 14} = \frac{1023 + 1722}{196} = \frac{2745}{196} \)

\( x = \frac{2745}{196} : \frac{5}{7} = \frac{2745}{196} \cdot \frac{7}{5} \)

\( x = \frac{2745 \cdot 7}{196 \cdot 5} = \frac{549 \cdot 1}{28 \cdot 1} = \frac{549}{28} \)

Ответ: А) x=0,7; Б) x=0,15; B) x=¹⁰/₇; Г) x=9; Д) x=⁶¹/₇₈₀; E) x=⁵⁴⁹/₂₈.