Question

Решите систему уравнений

Answer 1

Ответ:

$\left\{\begin{array}{l}3^{\frac{x-y}{2}}+3^{\frac{x-y}{4}}=12\\x^2+2y^2=3xy\end{array}\right\\\\\\a)\ \ }3^{\frac{x-y}{2}}+3^{\frac{x-y}{4}}=12\ \ ,\ \ \ \ (3^{x-y})^{\frac{1}{2}}+(3^{x-y})^{\frac{1}{4}}-12=0\ \ ,\\\\t=3^{\frac{x-y}{4}}>0\ \ ,\ \ \ t^2+t-12=0\ \ ,\ \ t_1=-4<0\ ,\ \ t_2=3\ \ (teorema\ Vieta)\\\\3^{\frac{x-y}{4}}=3^1\ \ ,\ \ \ \frac{x-y}{4}=1\ \ ,\ \ \ x-y=4\ \ ,\ \ \underline {x=y+4}$

$b)\ \ x^2+2y^2=3xy\ \ ,\ \ \ (y+4)^2+2y^2=3y(y+4)\ \ ,\\\\y^2+8y+16+2y^2=3y^2+12y\ \ ,\ \ \ 12y-8y=16\ \ ,\ \ 4y=16\ \ ,\ \ \underline {y=4}\\\\x=y+4=4+4=8$

$Otvet:\ (\, 8\, ;\, 4\, )\ .$

Answer 2

Ответ: (8;4)

Объяснение: