3) 1,9(y + 20)(20 - y) - 1,6 (y + 20)² = 116 - 3,5y² 4) 30(1,8- y)²+20(y +1,8)(y - 1,8) = 50y²+ 140,4. - id1716241320230308 от witerold 08.03.2023 19:51

Question

Алгебра: 3) 1,9(y + 20)(20 - y) - 1,6 (y + 20)² = 116 - 3,5y² 4) 30(1,8- y)²+20(y +1,8)(y - 1,8) = 50y²+ 140,4.​

witerold · 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

3) 1,9(y + 20)(20 - y) - 1,6 (y + 20)² = 116 - 3,5y² 4) 30(1,8- y)²+20(y +1,8)(y - 1,8) = 50y²+ 140,4.​

MOGZ ответил

3) 1,9(y + 20)(20 - y) - 1,6 (y + 20)² = 116 - 3,5y² 4) 30(1,8- y)²+20(y +1,8)(y - 1,8) = 50y²+ 140,4.