Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1329

$$ \text { a) }\left\{\begin{array}{l} 2 x^2+4 x y-5 y=1 \\ x^2+x y-6 y^2=0 \end{array}\right. $$
    x² + xy − 6y² = x² + 3xy − 2xy − 6y² = x(x + 3y) − 2y(x + 3y) = (x + 3y) = (x + 3y)(x − 2y)     (x + 3y)(x − 2y) = 0     $$ \left\{\begin{array} { l } { x + 3 y = 0 } \\ { 2 x ^ { 2 } + 4 x y - 5 y = 1 } \end{array} \left\{\begin{array}{l} x=-3 y \\ 2(-3 y)^2+4\lceil(-3 y) y-5 y=1 \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { x = - 3 y } \\ { 1 8 y ^ { 2 } - 1 2 y ^ { 2 } - 5 y - 1 = 0 } \end{array} \left\{\begin{array}{l} x=-3 y \\ 6 y^2-5 y-1=0 \end{array}\right.\right. $$
    6y² − 5y − 1 = 0     D = 5² − 4 · 6 · (−1) = 25 + 24 = 49
    y₁ = 5 + 7/12 = 12/12 = 1
    y₂ = 5 − 7/12 = −2/12 = −1/6
    $$ \left\{\begin{array} { l } { y _ { 1 } = 1 } \\ { x _ { 1 } = - 3 y = - 3 · 1= - 3 } \end{array} \left\{\begin{array}{l} y_1=1 \\ x_1=-3 \end{array}\right.\right. $$
    $$ \left\{\begin{array}{l} y_2=-\frac{1}{6} \\ x_2=-3 y=-3\left(-\frac{1}{6}\right)=\frac{1}{2}\left\{\begin{array}{l} y_2=-\frac{1}{6} \\ x_2=\frac{1}{2} \end{array}\right. \end{array}\right. $$
    $$ \left\{\begin{array}{l} y_3=\frac{5+\sqrt{89}}{32} \\ x_3=2 y=2 \frac{5+\sqrt{89}}{32}=\frac{5+\sqrt{89}}{16}\left\{\begin{array}{l} y_3=\frac{5+\sqrt{89}}{32} \\ x_3=\frac{5+\sqrt{89}}{16} \end{array}\right. \end{array}\right. $$
    $$ \left\{\begin{array}{l} y_4=\frac{5-\sqrt{89}}{32} \\ x_4=2 y=2 \frac{5-\sqrt{89}}{32}=\frac{5-\sqrt{89}}{16}\left\{\begin{array}{l} y_4=\frac{5-\sqrt{89}}{32} \\ x_4=\frac{5-\sqrt{89}}{16} \end{array}\right. \end{array}\right. $$
    (−3; 1), (1/2; 1/6), $$ \left(\frac{5+\sqrt{89}}{16} ; \frac{5+\sqrt{89}}{32}\right),\left(\frac{5-\sqrt{89}}{16} ; \frac{5-\sqrt{89}}{32}\right) $$
$$ \text { б) }\left\{\begin{array}{l} x(3 x-2 y)=y^2 \\ 3 y^2=2 x(x+2)-3 \end{array}\right. $$
    x(3x − 2y) − y² = 3x² − 2xy − y² = 3x² − 3xy + xy − y²     3x(x − y) + y(x − y) = (x − y)(3x + y)     (x − y)(3x + y) = 0
    $$ \left\{\begin{array}{l} x-y=0 \\ 3 y^2=2 x(x+2)-3 \end{array}\left\{\begin{array}{l} x=y \\ 3 y^2-2 y(y+2)+3=0 \end{array}\right\}\right. $$ $$ \left\{\begin{array} { l } { x = y } \\ { 3 y ^ { 2 } - 2 y ^ { 2 } - 4 y + 3 = 0 } \end{array} \left\{\begin{array}{l} x=y \\ y^2-4 y+3=0 \end{array}\right.\right. $$
    y² − 4y + 3 = 0     D = (−4)² − 4 · 1 · 3 = 16 − 12 = 4
    y₁ = 4 + 2/2 = 6/2 = 3
    y₂ = 4 − 2/2 = 2/2 = 1
    $$ \left\{\begin{array}{l} y_1=3 \\ x_1=3 \end{array}\right. $$
    $$ \left\{\begin{array}{l} y_2=1 \\ x_2=1 \end{array}\right. $$
    $$ \left\{\begin{array} { l } { 3 x + y = 0 } \\ { 3 y ^ { 2 } = 2 x ( x + 2 ) - 3 } \end{array} \left\{\begin{array}{l} y=-3 x \\ 3(-3 x)^2-2 x(x+2)+3=0 \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { y = - 3 x } \\ { 2 7 x ^ { 2 } - 2 x ^ { 2 } - 4 x + 3 = 0 } \end{array} \left\{\begin{array}{l} y=-3 x \\ 25 x^2-4 x+3=0 \end{array}\right.\right. $$
    25x² − 4x + 3 = 0     D = (−4)² − 4 · 25 · 3 = 16 − 300 = −284
    Нет корней     (3; 3), (1; 1)