Question

Задание на фото найти х/y можно, если будут найдены х и y
а можно ли найти х/y, не находя значения х и y?

Answer 1

х/у=(-3+-sqrt(5))/2

Объяснение:

Answer 2

$\left\{\begin{array}{l}x+y=3xy\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2=9(xy)^2\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}x^2+y^2+2xy=9(xy)^2\\x^2+y^2=\dfrac{1}{3}\end{array}\rightleft\{\begin{array}{l}\dfrac{1}{3}+2xy=9(xy)^2\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}9(xy)^2-2(xy)-\dfrac{1}{3}=0\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}27(xy)^2-6(xy)-1=0\\x^2+y^2=\dfrac{1}{3}\end{array}\right$

$\star \ \ t=xy\ ,\ 27t^2-6t-1=0\ ,\ \ D/4=9+27=36\ ,t_1=\dfrac{3-6}{27}=-\dfrac{1}{9}\ \ ,\ \ \ t_2=\dfrac{3+6}{27}=\dfrac{1}{3}\ \ \Rightarrow \ \ \ (xy)_1=-\dfrac{1}{9}\ \ ,\ \ (xy)_2=\dfrac{1}{3}\ \ \star$  

$a)\ \ \left\{\begin{array}{l}x+y=3xy\\xy=-\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}x+y=-\dfrac{1}{3}\\xy=-\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}y=-x-\dfrac{1}{3}\\x(x-\dfrac{1}{3} )=-\dfrac{1}{9}\end{array}\rightx^2-\dfrac{1}{3}\, x+\dfrac{1}{9}=0\ \ \Big|\cdot 9\ \ \ \Rightarrow \ \ \ 9x^2-3x+1=0\ \ ,\ \ D=9+36=45\ \ ,$

$x_1=\dfrac{-3-3\sqrt5}{2\cdot 9}=\dfrac{-1-\sqrt5}{6}\ \ ,\ \ \ x_2=\dfrac{-1+\sqrt5}{6}y_1=-x-\dfrac{1}{3}=\dfrac{1+\sqrt5}{6}-\dfrac{1}{3}=\dfrac{\sqrt5-1}{6}\ \ ,y_2=-x-\dfrac{1}{3}=\dfrac{1-\sqrt5}{6}-\dfrac{1}{3}=\dfrac{-\sqrt5-1}{6}boldsymbol{\dfrac{x_1}{y_1}}=\dfrac{-1-\sqrt5}{\sqrt5-1}=\dfrac{-(1+\sqrt5)^2}{(\sqrt5-1)(\sqrt5+1)}=\dfrac{-(6+2\sqrt5)}{5-1}=\bf \dfrac{-3-\sqrt5}{2}$

$\boldsymbol{\dfrac{x_2}{y_2}}=\dfrac{-1+\sqrt5}{-\sqrt5-1}=\dfrac{(\sqrt5-1)^2}{-(\sqrt5+1)(\sqrt5-1)}=\dfrac{6-2\sqrt5}{-(5-1)}=\dfrac{2(3-\sqrt5)}{-4}=\bf \dfrac{-3+\sqrt5}{2}$  

$b)\ \ \left\{\begin{array}{l}x+y=3xy\\xy=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}x+y=1\\xy=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}y=1-x\\x(1-x)=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}x+y=1\\x-x^2-\dfrac{1}{3}=0\end{array}\right-x^2+x-\dfrac{1}{3}=0\ \ \Big|\cdot (-3)\ \ \ \Rightarrow \ \ \ 3x^2-3x+1=0\ \ ,\ \ D=9-4\cdot 3=-3 < 0$

Так как  D<0 , то действительных корней нет . Система не имеет решений .

ответ:     $\boldsymbol{\dfrac{x_1}{y_1}=\dfrac{-3-\sqrt5}{2}}\ \ ,\ \ \boldsymbol{\dfrac{x_2}{y_2}=\dfrac{-3+\sqrt5}{2}}$   .