b6+b8=14
b10*b4=48
b6+b8=14
b1*q^9*b1*q3=48
b1^2*q*12=48
(b1*q6)^2=48
(b7)^2=48
Применим характеристическое свойство геометрической прогресии:
b6*b8=(b7)^2
b6*b8=48
b6=48/b8
b6+b8=14
48/b8 +b8=14
48+(b8)^2=14*b8
(b8)^2-14b8+48=0
b8(1)=6, b8(2)=8 (по теореме Виета)
b6=(1)=48/6=8, b6(2)=48/8=6
Ответ: 6;8 и 8;6
X = 6 + Y
(6+Y)Y = 16
6Y + Y^2 = 16
Y^2 + 6Y - 16 = 0
D = 36 - 4*(-16) = 36 + 64 = 100
Y 1 = -6 + 10 \2 = 4\2 = 2
Y2 = - 6 - 10 \ 2 = - 8
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X1 = 6 + Y1 = 6 + 2 = 8
X2 = 6 + Y2 = 6 - 8 = - 2
Решение
cosπ/12 - cos5π/12 = - 2*sin[(π/12 + 5π/12)/2]*sin[(π/12 - 5π/12)/2] =
= 2*sin(π/4)*sin(π/6) = 2*(√2/2)*(1/2) = √2/2