Решить систему 2^х-у = (1/4) ^-3/2; log2 x + log2 y = 2

Решение:

{2^х-у = (1/4) ^-3/2

{log2 x + log2 y = 2

используя тождество log a (x*y) = log a (x) + log a (y), получаем:

{2^х-у = (1/4) ^-3/2

{log2 (x * y) = 2

{2^х-у = (1/4) ^-3/2

{ x * y = (2) ^2

{2^х-у = (1/4) ^-3/2

{ x * y = 4

(1/4) ^-3/2 = 1 / (1/4) ^3/2 = 1/[1 / (4) ^3/2] = 1/[1/2^3] = 1/[1/8] = 8

{2^ (х-у) = 8

{ x * y = 4

{х-у=3

{ x * y = 4

{х=3+у

{ y = 4/x

{х=3+4/x

{ y = 4/x

{х^2-3x-4=0

{ y = 4/x

решаю квадратное уравнение:

D = (-3) ^2-4*1 * (-4) = 9+16 = 25

x1=[ - (-3) + √25] / (2*1) = (3+5) / 2 = 4

x2=[ - (-3) - √25] / (2*1) = (3-5) / 2 = - 1

далее:

{ x1 = 4

{ x2 = - 1

{ y = 4/x

{ x1 = 4

{ x2 = - 1

{ y1 = 4/4 = 1

{ y2 = 4/-1 = - 4

Ответ: x1=4, y1=1; x2=-1, y2=-4