Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1328

$$ \left\{\begin{array}{l} x y=-2 \\ (x-y)^2+x+y=10 \end{array}\right. $$
(x − y)² + x + y = 10 x² − 2xy + y² + x + y = 10 x² − 2 · (−2) + y² + x + y = 10 (x + y)² − 2xy + x + y + 4 − 10 = 0 (x + y)² − 2 · (−2) + x + y + 4 − 10 = 0 (x + y)² + (x + y) − 2 = 0 p = x + y p² + p − 2 = 0 D = 1 + 8 = 9
p₁ = −1 + 3/2 = 2/2 = 1
p₂ = −1 − 3/2 = −4/2 = −2
$$ \left\{\begin{array} { l } { p = - 2 } \\ { x + y = - 2 } \\ { x y = - 2 } \end{array} \left\{\begin{array} { l } { p = - 2 } \\ { y = - 2 - x } \\ { x ( - 2 - x ) = - 2 } \end{array} \left\{\begin{array}{l} p=-2 \\ y=-2-x \\ -x^2-2 x=-2 \end{array}\right.\right.\right. $$
x² + 2x − 2 = 0 D = 4 + 8 = 12
x₁ = $$ \frac{-2+2 \sqrt{3}}{2} $$ = −1 + $$\sqrt{3}$$
x₂ = $$ \frac{-2-2 \sqrt{3}}{2} $$ = −1 − $$\sqrt{3}$$
$$ \left\{\begin{array} { l } { x _ { 1 } = - 1 + \sqrt { 3 } } \\ { y _ { 1 } = - 2 - ( - 1 + \sqrt { 3 } ) } \end{array} \left\{\begin{array}{l} x_1=-1+\sqrt{3} \\ y_1=-1-\sqrt{3} \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = - 1 - \sqrt { 3 } } \\ { y _ { 2 } = - 2 - ( - 1 - \sqrt { 3 } ) } \end{array} \left\{\begin{array}{l} x_2=-1-\sqrt{3} \\ y_2=-1+\sqrt{3} \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { p = 1 } \\ { x + y = 1 } \\ { x y = - 2 } \end{array} \left\{\begin{array} { l } { p = 1 } \\ { y = 1 - x } \\ { x ( 1 - x ) = - 2 } \end{array} \left\{\begin{array}{l} p=1 \\ y=1-x \\ -x^2+x=-2 \end{array}\right.\right.\right. $$
x² − x − 2 = 0 D = 1 + 8 = 9
x₃ = 1 + 3/2 = 2
x₄ = 1 − 3/2 = −1
$$ \left\{\begin{array} { l } { x _ { 3 } = 2 } \\ { y _ { 3 } = 1 - 2 = - 1 } \end{array} \left\{\begin{array}{l} x_3=2 \\ y_3=-1 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} x_4=-1 \\ y_4=1+1 \end{array}=2\left\{\begin{array}{l} x_4=-1 \\ y_4=2 \end{array}\right.\right. $$
(−1 + $$\sqrt{3}$$; −1 − $$\sqrt{3}$$), (−1 − $$\sqrt{3}$$; −1 + $$\sqrt{3}$$), (2; −1), (−1; 2)