Question

решите систему уравнений​ (методом замены)

Answer 1

Ответ:

$\left\{\begin{array}{l}x+y+\dfrac{x^2}{y^2}=7\ ,\\\dfrac{(x+y)\, x^2}{y^2}=12\end{array}\right\ \ \ zamena:\ \left\{\begin{array}{l}x+y=u\\\dfrac{x^2}{y^2}=v\\y\ne 0\end{array}\right\ \ \left\{\begin{array}{l}u+v=7\\uv=12\end{array}\right\\\\\\ \left\{\begin{array}{l}u=7-v\\(7-v)\, v=12\end{array}\right\ \ \left\{\begin{array}{l}u=7-v\\v^2-7v+12=0\end{array}\right\ \ \left\{\begin{array}{l}u=7-v\\v_1=3\ ,\ v_2=4\ (teor.Vieta)\end{array}\right$

$\left\{\begin{array}{l}u_1=4\ ,\ u_2=3\\v_1=3\ ,\ v_2=4\end{array}\right\\\\\\a)\ \ \left\{\begin{array}{l}x+y=4\\\dfrac{x^2}{y^2}=3\end{array}\right\ \ \left\{\begin{array}{l}x=4-y\\16-8y+y^2=3y^2\end{array}\right\ \ \left\{\begin{array}{l}x=4-y\\2y^2+8y-16=0\end{array}\right$

$\left\{\begin{array}{l}x=4-y\\y^2+4y-8=0\end{array}\right\ \ \left\{\begin{array}{l}x_1=6+2\sqrt3\ \ \ ,\ x_2=6-2\sqrt3\\y_1=-2-2\sqrt3\ ,\ y_2=-2+2\sqrt3\end{array}\right\\\\\\\star \ y^2+4y-8=0\ \ ,\ \ D/4=2^2+8=12\ ,\ y_1=-2-2\sqrt3\ ,\ y_2=-2+\sqrt3\ \star$

$b)\ \ \left\{\begin{array}{l}x+y=3\\\dfrac{x^2}{y^2}=4\end{array}\right\ \ \left\{\begin{array}{l}x=3-y\\9-6y+y^2=4y^2\end{array}\right\ \ \left\{\begin{array}{l}x=3-y\\3y^2+6y-9=0\end{array}\right\\\\\\\left\{\begin{array}{l}x=3-y\\y^2+2y-3=0\end{array}\right\ \ \left\{\begin{array}{l}x_1=6\ ,\ x_2=2\\y_1=-3\ ,\ y_2=1\ (teor.\ Vieta)\end{array}\right$

$Otvet:\ \ (\ 6+2\sqrt3\ ;-2-2\sqrt3\ )\ ,\ (\ 6-2\sqrt3\ ;-2+2\sqrt3\ )\ ,\ (\ 6\ ;-3\ )\ ,\ (\ 2\ ;\ 1\ )\ .$