Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1327

а) $$ \left\{\begin{array}{l} 7 x y+2 x^2-4 y^2=0 \\ x^2-5 x y+y=-11 \end{array}\right. $$
    7xy + 2x² − 4y² = 0 | : y²     2(x/y)² + 7(x/y) − 4 = 0     p = x/y     2p² + 7p − 4 = 0     D = 7² + 4 · 2 · 4 = 49 + 32 = 81
    p₁ = −7 + 9/4 = 2/4 = 0,5
    p₂ = −7 − 9/4 = −16/4 = −4
    x/y = 0,5 ⇒ y = 2x     x/y = −4 ⇒ y = −0,25x     x² − 5xy + y = −11     x² + y(1 − 5x) = −11     y(1 − 5x) = −11 − x²
    y = −11 − x²/1 − 5x     y = 11 + x²/5x − 1     y = 2x     11 + x²/5x − 1 = 2x     2x(5x − 1) = 11 + x²     10x² − 2x − 11 − x² = 0     9x² − 2x − 11 = 0     D = 4 + 396 = 400     x₁ = 2 + 20/18 = 22/18 = 11/9     x₂ = 2 − 20/18 = −18/18 = −1     y = −0,25x
    11 + x²/5x − 1 = −0,25x     −0,25x(5x − 1) = 11 + x²     −1,25x² + 0,25x − 11 − x² = 0     −2,25x² + 0,25x − 11 = 0 | : (−0,25)     9x² − x + 44 = 0     D = 1 − 1584 = −1583
    Нет решений
    $$ \left\{\begin{array}{l} x_1=\frac{11}{9} \\ y=\frac{11+x^2}{5 x-1}=2 \frac{4}{9} \end{array}\right. $$
    $$ \frac{11+\left(\frac{11}{9}\right)^2}{5 \frac{11}{9}-1}=\frac{11+\frac{121}{81}}{\frac{55}{9}-1}=\frac{11+1 \frac{40}{81}}{6 \frac{1}{9}-1}=\frac{12 \frac{40}{81}}{5 \frac{1}{9}} $$ = 1012/81 : 46/9 = 1012/81 · 9/46 = 22/9 = 24/9
    $$ \left\{\begin{array}{l} x_2=-1 \\ y=\frac{11+x^2}{5 x-1}=-2 \end{array}\right. $$
    11 + (−1)²/5 · (−1) − 1 = 11 + 1/−5 − 1 = 12/−6 = −2
    (11/9; 24/9); (−1; −2)
$$ \text { б) }\left\{\begin{array}{l} 6 x^2+2 x y-3 x-y=0 \\ 2 x^2-y^2+2 x+y=\frac{3}{2} \end{array}\right. $$
    $$ \left[\begin{array}{l} \left\{\begin{array} { l } { x = 0 , 5 } \\ { 2 x ^ { 2 } - y ^ { 2 } + 2 x + y = \frac { 3 } { 2 } } \end{array} \left[\begin{array}{l} x=0,5 \\ 0,5-y^2+1+y=\frac{3}{2} \end{array}\right.\right. \\ \left\{\begin{array} { l } { y = - 3 x } \\ { 2 x ^ { 2 } - y ^ { 2 } + 2 x + y = \frac { 3 } { 2 } } \end{array} \left[\begin{array}{l} y=-3 x \\ 2 x^2-9 x^2+2 x-3 x=\frac{3}{2} \end{array}\right.\right. \end{array}\right. $$ $$ \left[\begin{array}{l} \left\{\begin{array}{l} x=0,5 \\ 0,5-y^2+1+y=\frac{3}{2} \end{array}\right. \\ \left\{\begin{array}{l} y=-3 x \\ 2 x^2-9 x^2+2 x-3 x=\frac{3}{2} \end{array}\right. \end{array}\right. $$
    $$ \left[\begin{array} { l } { \{ \begin{array} { l } { x = 0 , 5 } \\ { y - y ^ { 2 } = 0 } \end{array} } \\ { \{ \begin{array} { l } { y = - 3 x } \\ { 7 x ^ { 2 } + x + 1 , 5 = 0 } \end{array} } \end{array} \left[\begin{array}{l} \left\{\begin{array} { l } { x = 0 , 5 } \\ { y ( 1 - y ) = 0 } \\ { \{ \begin{array} { l } { y = - 3 x } \\ { \varnothing } \end{array} } \end{array} \Rightarrow \left\{\begin{array}{l} x=0,5 \\ y_1=0 \\ y_2=1 \end{array}\right.\right. \end{array}\right.\right. $$
    7x² + x + 1,5 = 0     D = 1 − 42 = −41     (0,5; 0), (0,5; 1)