Question

X^2-3xy+2y^2=0 , x^2+y^2=20 система!

Answer 1

$\left\{ \begin{array}{l} x^2 - 3xy + 2y^2 = 0 ; \\ x^2 + y^2 = 20 . \end{array}$

Решение:

При $y = 0$ и $x ,$ судя по первому уравнению, тоже должно быть равно нулю, а это не подходит во второе уравнение, значит $y \neq 0$

$\left\{ \begin{array}{l} x^2 - 3xy + 2y^2 = 0 || : y^2 ; \\ x^2 + y^2 = 20 . \end{array}$

$\left\{ \begin{array}{l} \frac{x^2}{y^2} - 3 \frac{xy}{y^2} + 2 \frac{y^2}{y^2} = \frac{0}{y^2} ; \\ \\ x^2 + y^2 = 20 . \end{array}$

$\left\{ \begin{array}{l} ( \frac{x}{y} )^2 - 3 \frac{x}{y} + 2 = 0 ; \Rightarrow D = 3^2 - 4 \cdot 2 = 1^2 ; \Rightarrow ( \frac{x}{y} )_{12} = \frac{ 3 - 1 }{2} = 1 ; ( \frac{x}{y} )_{34} = \frac{ 3 + 1 }{2} = 2 ; \\ \\ x^2 + y^2 = 20 . \end{array}$

$\left\{ \begin{array}{l} \left[ \begin{array}{l} x_{12} = y_{12} ; \\ x_{34} = 2 y_{34} ; \end{array} \\ \\ x^2 + y^2 = 20 . \end{array}$

$\left[ \begin{array}{l} \left\{ \begin{array}{l} x_{12} = y_{12} ; \\ y_{12}^2 + y_{12}^2 = 20 ; \end{array} \\ \\ \left\{ \begin{array}{l} x_{34} = 2 y_{34} ; \\ ( 2 y_{34} )^2 + y_{34}^2 = 20 . \end{array} \end{array}$

$\left[ \begin{array}{l} \left\{ \begin{array}{l} x_{12} = y_{12} ; \\ 2 y_{12}^2 = 20 ; \end{array} \\ \\ \left\{ \begin{array}{l} x_{34} = 2 y_{34} ; \\ 4 y_{34}^2 + y_{34}^2 = 20 . \end{array} \end{array}$

$\left[ \begin{array}{l} \left\{ \begin{array}{l} x_{12} = y_{12} ; \\ y_{12}^2 = 10 ; \end{array} \\ \\ \left\{ \begin{array}{l} x_{34} = 2 y_{34} ; \\ 5 y_{34}^2 = 20 . \end{array} \end{array}$

$\left[ \begin{array}{l} x_1 = y_1 = -\sqrt{10} ; \\ x_2 = y_2 = \sqrt{10} ; \\ \\ \left\{ \begin{array}{l} x_{34} = 2 y_{34} ; \\ y_{34}^2 = 4 . \end{array} \end{array}$

$\left[ \begin{array}{l} ( x_1 ; y_1 ) = ( -\sqrt{10} ; -\sqrt{10} ) ; \\ ( x_2 ; y_2 ) = ( \sqrt{10} ; \sqrt{10} ) ; \\ \left\{ \begin{array}{l} y_3 = -2 ; \\ x_3 = -4 ; \end{array} \\ \left\{ \begin{array}{l} y_4 = 2 ; \\ x_4 = 4 . \end{array} \end{array}$

О т в е т :
$( x_1 ; y_1 ) = ( -\sqrt{10} ; -\sqrt{10} )$ .
$( x_2 ; y_2 ) = ( \sqrt{10} ; \sqrt{10} )$ .
$( x_3 ; y_3 ) = ( -2 ; -4 )$ .
$( x_4 ; y_4 ) = ( 2 ; 4 )$ .

или:

О т в е т : $( x ; y ) \in \{ ( -\sqrt{10} ; -\sqrt{10} ) , ( \sqrt{10} ; \sqrt{10} ) , ( -4 , -2 ) , ( 4 , 2 ) \}$ ;

или:

О т в е т : $( x ; y ) \in \{ ( \pm \sqrt{10} ; \pm \sqrt{10} ) , ( \pm 4 , \pm 2 ) \} .$