найдите такие значения переменной x, при которых числа 4,x√2,8 образуют геометрическую прогрессию.
b1 = 4; b2 = b1*q = 4q = x√2; b3 = b1*q^2 = 4q^2 = 8
q^2 = 8/4 = 2
q1 = -√2; x1 = 4q/√2 = -4√2/√2 = -4
q2 = √2; x2 = 4q/√2 = 4√2/√2 = 4
Ответ: x1 = -4; x2 = 4
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b1 = 4; b2 = b1*q = 4q = x√2; b3 = b1*q^2 = 4q^2 = 8
q^2 = 8/4 = 2
q1 = -√2; x1 = 4q/√2 = -4√2/√2 = -4
q2 = √2; x2 = 4q/√2 = 4√2/√2 = 4
Ответ: x1 = -4; x2 = 4