Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 710

а) $$ \left\{\begin{array}{l} x^2+y^2+3 x y=-1\left\{\begin{array}{l} (-2 y)^2+y^2+3(-2 y) · y=-1 \\ x+2 y=0 \end{array}\right. \\ x=-2 y \end{array}\right. $$ $$ \left\{\begin{array}{l} 4 y^2+y^2-6 y^2+1=0 \\ x=-2 y \end{array}\right. $$
$$ \left\{\begin{array} { l } { - y ^ { 2 } = - 1 } \\ { x = - 2 y } \end{array} \left\{\begin{array} { l } { y ^ { 2 } = 1 } \\ { x = - 2 y } \end{array} \left\{\begin{array}{l} y= \pm 1 \\ x=-2 y \end{array}\right.\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 1 } = 1 } \\ { x _ { 1 } = - 2 · 1 = - 2 } \end{array} \left\{\begin{array}{l} y_1=1 \\ x_1=-2 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 1 } \\ { x _ { 2 } = - 2 ( - 1 ) = - 2 } \end{array} \left\{\begin{array}{l} y_2=-1 \\ x_2=2 \end{array}\right.\right. $$
(−2; 1), (2; −1)
б) $$ \left\{\begin{array} { l } { u + 2 v = 4 } \\ { u ^ { 2 } + u v - v = - 5 } \end{array} \left\{\begin{array}{l} u=4-2 v \\ (4-2 v)^2+(4-2 v) v-v=-5 \end{array}\right.\right. $$ $$ \left\{\begin{array}{l} u=4-2 v \\ 16-16 v+4 v^2+4 v-2 v^2-v+5=0 \end{array}\right. $$
$$ \left\{\begin{array}{l} u=4-2 v \\ 2 v^2-13 v+21=0 \end{array}\right. $$
2v² − 13v + 21 = 0 D = b² − 4ac = (−13)² − 4 · 2 · 21 = 169 − 168 = 1 > 0, имеет 2 корня
v₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{13+\sqrt{1}}{2 · 2} $$ = 13 + 1/4 = 14/4 = 7/2 = 3,5
v₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{13-\sqrt{1}}{2 · 2} $$ = 13 − 1/4 = 12/4 = 3
v₁ = 3,5, v₂ = 3
$$ \left\{\begin{array} { l } { v _ { 1 } = 3 , 5 } \\ { u _ { 1 } = 4 - 2 · 3 , 5 = - 3 } \end{array} \left\{\begin{array}{l} v_1=3,5 \\ u_1=-3 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { v _ { 2 } = 3 } \\ { u _ { 2 } = 4 - 2 · 3 = - 2 } \end{array} \left\{\begin{array}{l} v_2=3 \\ u_2=-2 \end{array}\right.\right. $$