Question

помогите решить систему
2^x+2^y=12
2^x+y=32

Answer 1

$left { {{2^{x}+2^{y}=12} atop {2^{x+y}=32}} right. left { {{2^{x}+2^{y}=12} atop {2^{x+y}=2^{5}}} right. left { {{2^{x}+2^{y}=12} atop {x+y = 5}} right. left { {{2^{x}+2^{y}=12} atop {x=5-y}} right. \ \ 2^{5-y} + 2^{y} = 12 (2^{5}/2^{y}) + 2^{y} = 12 | *2^y 2^5 + 2^{2y} = 12*2^y\\ 2^y = a\\32 + a^2 = 12a\a^2 - 12a +32 =0\\a_1 + a_2 = 12\a_1 * a_2 = 32\\a_1 = 4\a_2 =8\\2^{y_1} = 4\y_1 = 2\\2^{y_2} = 8\y_2 = 3\\x_1 = 5 - y_1 = 5 -2 = 3\\x_2 = 5-y_2=5 - 3 = 2$

Ответ: x_1 = 3, y_1 = 2; x_2 = 2, y_2 = 3.

Answer 2

госпадя

Answer 3

Погоди инутку

Answer 4

кароч с ответом сам разберёшься... )))