Question

решите систему уравнений рациональным способом ​

Answer 1

Ответ:  
( 2 ; -1 ) ,  (-2 ; 1  ) , ( 1 ; -2 )  ,  (-1 ; 2  )

Объяснение:

$\ominus \left\{ \begin{array}{l} x^2+y^2=5 \\\\ x^2+y^2+3xy=-1 \end{array}\right. \Leftrightarrow \\\\\\ -3xy=6 \\\\xy=-2$

$x^2+y^2+\boldsymbol{2xy} =(x+y)^2 =5-2\cdot 2=1 \\\\x^2+y^2-\boldsymbol{2xy} =(x-y)^2 =5-2\cdot (-2)=9$

$\left\{ \begin{array}{l} (x+y)^2=1 \\\\ (x-y)^2=9 \end{array}\right. \Leftrightarrow$

1 = 1·1

1=(-1)·(-1)

9 = 3·3

9 = (-3)·(-3)

$\left [ \begin{array}{l} \left\{ \begin{array}{l} x+y=1 \\\\ x-y=3 \\ \end{array}\right. \Leftrightarrow x_1=2 ~~ ; ~~y_1=-1 \\\\ \left\{ \begin{array}{l} x+y=-1 \\\\ x-y=-3 \end{array}\right. \Leftrightarrow x_2=-2 ~~ ; ~~y_2=1 \end{array}\right.$



$\left [ \begin{array}{l} \left\{ \begin{array}{l} x+y=-1 \\\\ x-y=3 \\ \end{array}\right. \Leftrightarrow x_3=1 ~~ ; ~~y_3=-2 \\\\ \left\{ \begin{array}{l} x+y=1 \\\\ x-y=-3 \end{array}\right. \Leftrightarrow x_4=-1 ~~ ; ~~y_4=2 \end{array}\right.$