Question

Решите систему уравнений:

Answer 1

Поработаем со вторым

x² + 2y² = 3xy  

x² - 2xy + y² + y² - xy = 0  

(x - y)² - y(x - y) = 0  

(x - y)(x - 2y) = 0  

Теперь подставляем в первое  

1). x = y  

3^(x - x)/2 + 3^(x - x)/4 = 3^0 + 3^0 = 2  

2 ≠ 12  

Нет решений  

2). x = 2y  

3^(2y - y)/2 + 3^(2y - y)/4 = 12  

3^y/2 + 3^y/4 = 12  

3^y/4 = z > 0  

z² + z = 12  

D = 1 + 48 = 49  

z12 = (-1 +- 7)/2 = 3  -4  

z = -4 < 0 не корень  

z = 3  

3^y/4 = 3  

y/4 = 1  

y = 4  

x = 2y = 8  

Ответ (8, 4)

Answer 2

$1)3^{\dfrac{x-y}{2} }+3^{\dfrac{x-y}{4} }=12\\\\\\3^{\dfrac{x-y}{4} }=m \ , \ m>0 \ \Rightarrow \ 3^{\dfrac{x-y}{2} }=m^{2} \\\\m^{2} +m-12=0\\\\Teorema \ Vieta:\\\\m_{1} =3\\\\m_{2} =-4<0-neyd\\\\3^{\dfrac{x-y}{4} } =3\\\\\frac{x-y}{4} =1\\\\x-y=4\\\\\\\left\{\begin{array}{l}x-y=4\\x^{2}+2y^{2}=3xy \end{array}\right\\\\\\\left\{\begin{array}{l}x=4+y\\x^{2}+2y^{2}=3xy \end{array}\right$

$\left\{\begin{array}{l}x=4+y\\(4+y)^{2}+2y^{2}=3\cdot(4+y)\cdot y \end{array}\right \\\\\\\left\{\begin{array}{l}x=4+y\\16+8y+y^{2}+2y^{2}-12y-3y^{2} =0\end{array}\right \\\\\\\left\{\begin{array}{l}x=4+y\\4y=16 \end{array}\right \\\\\\\left\{\begin{array}{l}x=4+y\\y=4 \end{array}\right\\\\\\\left\{\begin{array}{l}x=8\\y=4 \end{array}\right\\\\\\Otvet:{(8 \ ; \ 4)}$