Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1326

$$ \text { a) }\left\{\begin{array}{l} x^2-3 y^2-y=-6 \\ 2 x^2-3 y^2=-4 \end{array}\right. $$
    $$ \left\{\begin{array} { l } { x ^ { 2 } - 3 y ^ { 2 } - y = - 6 } \\ { 2 x ^ { 2 } - 3 y ^ { 2 } = - 4 } \end{array} \left\{\begin{array} { l } { - x ^ { 2 } - y = - 4 + 6 } \\ { 2 x ^ { 2 } - 3 y ^ { 2 } = - 4 } \end{array} \left\{\begin{array}{l} y=-x^2+2 \\ 2 x^2-3\left(-x^2+2\right)^2=-4 \end{array}\right.\right.\right. $$
    2x² − 3x⁴ + 12x² − 12 + 4 = 0     −3x⁴ + 14x² − 8 = 0     3x⁴ − 14x² + 8 = 0     Обозначим     p = x²     3p² − 14p + 8 = 0     D = 196 − 4 · 3 · 8 = 196 − 96 = 100
    p₁ = 14 + 10/6 = 24/6 = 4
    p₂ = 14 − 10/6 = 4/6 = 2/3
    x² = 4     x = ±2     x² = 2/3     x = $$ \pm \sqrt{\frac{2}{3}} $$
    $$ \left\{\begin{array} { l } { x = \pm 2 } \\ { y = - 4 + 2 = - 2 } \end{array} \left\{\begin{array}{l} x= \pm 2 \\ y=-2 \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { x = \pm \sqrt { \frac { 2 } { 3 } } } \\ { y = - \frac { 2 } { 3 } + 2 = 1 \frac { 1 } { 3 } } \end{array} \left\{\begin{array}{l} x= \pm \sqrt{\frac{2}{3}} \\ y=1 \frac{1}{3} \end{array}\right.\right. $$
    (2; −2), (−2; −2), (2/3; 11/3), (−2/3; 11/3)
$$ \text { б) }\left\{\begin{array}{l} 2 x^2+x y=16 \\ 3 x^2+x y-x=18 \end{array}\right. $$
    $$ \left\{\begin{array} { l } { 2 x ^ { 2 } + x y = 1 6 } \\ { 3 x ^ { 2 } + x y - x = 1 8 } \end{array} \left\{\begin{array} { l } { x ^ { 2 } - x = 2 } \\ { 2 x ^ { 2 } + x y = 1 6 } \end{array} \left\{\begin{array}{l} x^2-x-2=0 \\ 2 x^2+x y=16 \end{array}\right.\right.\right. $$
    D = 1 + 4 · 1 · 2 = 1 + 8 = 9
    x₁ = 1 + 3/2 = 4/2 = 2
    x₂ = 1 − 3/2 = −2/2 = −1
    $$ \left\{\begin{array} { l } { x = 2 } \\ { 2 · 2 ^ { 2 } + 2 y = 1 6 } \end{array} \left\{\begin{array} { l } { x = 2 } \\ { y = \frac { 1 6 - 8 } { 2 } = 4 } \end{array} \left\{\begin{array}{l} x=2 \\ y=4 \end{array}\right.\right.\right. $$
    $$ \left\{\begin{array} { l } { x = - 1 } \\ { 2 ( - 1 ) ^ { 2 } - y = 1 6 } \end{array} \left\{\begin{array} { l } { x = - 1 } \\ { y = 2 - 1 6 = - 1 4 } \end{array} \left\{\begin{array}{l} x=-1 \\ y=-14 \end{array}\right.\right.\right. $$
    (2; 4), (−1; −14)