Алексей
Ев
Евгений
Ответ
{ 2x - y = 12
{ log2 x - 3log2 y = -3 => log2 x - log2 y^3 = -3
ОДЗ: x>0 и y>0
{ y = 2x - 12
{ log2 x/y^3 = -3 => x/y^3 = 2^(-3)
{ y = 2x - 12
{ x/(2x-12)^3 = 1/2^3 => x/{2*(x-6)}^3 = 1/2^3
{ y = 2x - 12
{ x = 2^3*(x-6)^3/2^3 => x = (x-6)^3
{ y = 2*(x - 6) => x-6 = y/2
{ x = (x-6)^3
{ x - 6 = y/2 => x = y/2 + 6
{ x = (y/2)^3
y/2 + 6 = (y/2)^3
y/2 = t
t^3 - t - 6 = 0
(t^3 - - (t - 2) = 0
(t-2)(t^2 + 2t + 4) - (t - 2) = 0
(t - 2)(t + 2)^2 = 0
t1 = 2 => y/2=2 =>
y=4 => x = (y/2)^3 = (4/2)^3 = 8
t2 = -2 => y/2=-2 =>
y=-4 => x = (y/2)^3 = (-4/2)^3 = -8 (по ОДЗ не подходит)
=> ответ: x=8, y=4
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