Упр.103 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)

1) 140 — (x : 7 + 29) · 4 = 12; 3) 100 : [19 + (15х — 84) : 6] = 4;
2) 720 : (5х — 12) — 56 = 34; 4) [72 — 64 : (40 — 8x)] · 4 = 272.

1) $$140 — (x : 7 + 29)\cdot 4 = 12$$

$$ (x : 7 + 29)\cdot 4 = 140 — 12 \\ (x : 7 + 29)\cdot 4 = 128 \\ x : 7 + 29 = 128 : 4 \\ x : 7 + 29 = 32 \\ x : 7 = 32 — 29 \\ x : 7 = 3 \\ x = 3 \cdot 7 \\ x = 21 $$

2) $$720 : (5x — 12) — 56 = 34$$

$$ 720 : (5x — 12) = 34 + 56 \\ 720 : (5x — 12) = 90 \\ 5x — 12 = 720 : 90 \\ 5x — 12 = 8 \\ 5x = 8 + 12 \\ 5x = 20 \\ x = 20 : 5 \\ x = 4 $$

3) $$100 : [19 + (15x — 84) : 6] = 4$$

$$ 19 + (15x — 84) : 6 = 100 : 4 \\ 19 + (15x — 84) : 6 = 25 \\ (15x — 84) : 6 = 25 — 19 \\ (15x — 84) : 6 = 6 \\ 15x — 84 = 6 \cdot 6 \\ 15x — 84 = 36 \\ 15x = 36 + 84 \\ 15x = 120 \\ x = 120 : 15 \\ x = 8 $$

4) $$[72 — 64 : (40 — 8x)]\cdot 4 = 272$$

$$ 72 — 64 : (40 — 8x) = 272 : 4 \\ 72 — 64 : (40 — 8x) = 68 \\ 64 : (40 — 8x) = 72 — 68 \\ 64 : (40 — 8x) = 4 \\ 40 — 8x = 64 : 4 \\ 40 — 8x = 16 \\ 8x = 40 — 16 \\ 8x = 24 \\ x = 24 : 8 \\ x = 3 $$

Ответ

1) $$x = 21$$; 2) $$x = 4$$; 3) $$x = 8$$; 4) $$x = 3$$.