в заданной прогрессии 6 членов
Объяснение:
1. Для заданной геометрической прогрессии B(n) известно следующее:
B1 + Bn = 66;
B1 = 66 - Bn;
2. B2 * B(n - 1) = 128;
(B1 * q) * (B1 * q^(n - 2) = B1 * (B1 * q* q^(n - 2)) =
B1 * (B1 * q^(n - 1)) = B1 * Bn = 128;
(66 - Bn) * Bn = 128;
Bn² - 66 * Bn + 128 = 0;
Bn1,2 = 33 +- sqrt(33² - 128) = 33 +- 31;
Bn = 33 + 31 = 64 (прогрессия возрастающая);
B1 = 66 - Bn = 66 - 64 = 2;
3. Вычислим n:
B1 * Bn = B1² * q^(n - 1) = 128;
q^(n - 1) = 128 / B1² = 128 / 2² = 32 = 2^5;
n - 1 = 5;
n = 5 + 1 = 6.
1.
а) 3b+(5a–7b) = 3b+5a–7b = 5a–4b
б) –(8c–4) +4 = –8c+4+4 = 8–8c
в) (2+3x) +(7x–2) = 2+3x+7x–2 = 10x
г) 3(8m–4)+6m = 3×8m–3×4+6m=24m–12+6m=30m–12
д) 15–5(1–a)–6a = 15–5–5a–6a= 10–11a
е) (2a–7y)–(5a–7) = 2a–7y–5a+7 = –3a–7y±7
ж) 14b–(15b+y)–(y+10b) = 14b–15b–y–y–10b = –11b–2y
з) 7(5a+8)–11a–58 = 7×5a+7×8–11a–58 = 35a+56–11a–58 = 24a–2
и) 9x+3(15–8x)–35 = 9x+3×15–3×8x–35 = 9x+45–24x–35 = 10–15x
к) 33–8(11b–1) –2b = 33–8×11b–8–2b = 33–88b–8–2b = 25–90b
2.
а) 0,7b+0,3(b–5) = 0,7b+0,3b–0,3×5 = b–1,5 = –0,81–1,5 = –2,31
б) (y–7)–(14–y) = y–7–14+y = 2y–21 = –0,6–21= –21,6
Объяснение:
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