15 51.1) 16,05x +1,8x = 3,63 - 0,3x; 3) 1,09 +5,8y = 38,29 + 15,1y; 5 5) x+2 3 = 3 28 4 X; TT - id1365240720220906 от лиза2677 06.09.2022 17:48

Question

Алгебра: 15 51.1) 16,05x +1,8x = 3,63 - 0,3x; 3) 1,09 +5,8y = 38,29 + 15,1y; 5 5) x+2 3 = 3 28 4 X; TT​

лиза2677 · Accepted Answer

B1 + b1q^3 = -49
b1q + b1q^2 = 14 разделим первое уравнение на 2-е
(1 + q^3)/(q +q^2) = -7/2
(1+q)(1 -q +q^2)/q(1 +q) = -7/2
(1 -q +q^2) /q = -7/2
2(1 - q +q^2) = -7q
2 -2q +2q^2 +7q = 0
2q^2 +5q +2 = 0
D = b^2 -4ac = 25 -16 = 9
q1= -1/2, a) b1 + b1q^3 = -49    б) q2 =-2     b1 + b1q^3 = -49
b1 +b1*(-1/8) = -49    b1 + b1*(-8) = -49
7/8 b1 = -49    -7b1 = -49
b1 = -49: 7/8= -49*8/7= =56    b1 = 7

15 51.1) 16,05x +1,8x = 3,63 - 0,3x;3) 1,09 +5,8y = 38,29 + 15,1y;55) x+23= 3284X;TT​

MOGZ ответил

15 51.1) 16,05x +1,8x = 3,63 - 0,3x;
3) 1,09 +5,8y = 38,29 + 15,1y;
5
5) x+2
3
= 3
28
4
X;
TT