Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 1325

а) $$ \left\{\begin{array}{l} \frac{2}{y-1}+\frac{3}{x+1}=\frac{5}{2} \\ \frac{1}{x-2}=-\frac{3}{y} \end{array}\right. $$
    ОДЗ: x ≠ −1; 2, y ≠ 0; 1
    $$ \left\{\begin{array} { l } { \frac { 2 } { y - 1 } + \frac { 3 } { x + 1 } = \frac { 5 } { 2 } } \\ { \frac { 1 } { x - 2 } = - \frac { 3 } { y } } \end{array} \left\{\begin{array}{l} \frac{2}{y-1}+\frac{3}{x+1}=\frac{5}{2} \\ y=-3(x-2)=-3 x+6 \end{array}\right.\right. $$ $$ \left\{\begin{array}{l} \frac{2}{-3 x+6-1}+\frac{3}{x+1}=\frac{5}{2}\left\{\begin{array}{l} \frac{2}{-3 x+5}+\frac{3}{x+1}=\frac{5}{2} \\ y=-3 x+6 \end{array}\right. \\ y=-3 x+6 \end{array}\right. $$
    2(x + 1) + 3(−3x + 5)/(−3x + 5)(x + 1) = 5/2
    2x + 2 − 9x + 15/(−3x + 5)(x + 1) = 5/2
    −7x + 17/(−3x + 5)(x + 1) = 5/2
    2(−7x + 17) = 5(−3x² − 3x + 5x + 5)     −14x + 34 + 15x² − 10x − 25 = 0     15x² − 24x + 9 = 0     5x² − 8x + 3 = 0     D = 64 − 4 · 5 · 3 = 64 − 60 = 4
    x₁ = 8 + 2/10 = 10/10 = 1
    x₂ = 8 − 2/10 = 6/10 = 0,6
    $$ \left\{\begin{array} { l } { x _ { 1 } = 1 } \\ { y _ { 1 } = - 3 · 1 + 6 = 3 } \end{array} \left\{\begin{array}{l} x_1=1 \\ y_1=3 \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { x _ { 2 } = 0 , 6 } \\ { y _ { 2 } = - 3 · 0 , 6 + 6 = 4 , 2 } \end{array} \left\{\begin{array}{l} x_2=0,6 \\ y_2=4,2 \end{array}\right.\right. $$
    (1; 3), (0,6; 4,2)
б) $$ \left\{\begin{array}{l} \frac{1}{y+1}=\frac{2}{x-1} \\ \frac{4}{x+2}+\frac{1}{y-1}=\frac{1}{3} \end{array}\right. $$
    ОДЗ: x ≠ −1; 2, y ≠ −1; 1
    $$ \left\{\begin{array} { l } { \frac { 1 } { y + 1 } = \frac { 2 } { x - 1 } } \\ { \frac { 4 } { x + 2 } + \frac { 1 } { y - 1 } = \frac { 1 } { 3 } } \end{array} \left\{\begin{array} { l } { x - 1 = 2 ( y + 1 ) } \\ { \frac { 4 } { x + 2 } + \frac { 1 } { y - 1 } = \frac { 1 } { 3 } } \end{array} \left\{\begin{array}{l} x=2 y+2+1 \\ \frac{4}{x+2}+\frac{1}{y-1}=\frac{1}{3} \end{array}\right.\right.\right. $$
    $$ \left\{\begin{array} { l } { x = 2 y + 3 } \\ { \frac { 4 } { 2 y + 3 + 2 } + \frac { 1 } { y - 1 } = \frac { 1 } { 3 } } \end{array} \left\{\begin{array}{l} x=2 y+3 \\ \frac{4}{2 y+5}+\frac{1}{y-1}=\frac{1}{3} \end{array}\right.\right. $$
    4(y − 1) + 2y + 5/(2y + 5)(y − 1) = 1/3
    4y − 4 + 2y + 5/(2y + 5)(y − 1) = 1/3
    6y + 1/(2y + 5)(y − 1) = 1/3
    3(6y + 1) = (2y + 5)(y − 1)     18y + 3 = 2y² − 2y + 5y − 5     18y + 3 − 2y² − 3y + 5 = 0     −2y² + 15y + 8 = 0     D = 225 + 4 · 2 · 8 = 225 + 64 = 289
    y₁ = −15 + 17/−4 = 2/−4 = −0,5
    y₂ = −15 − 17/−4 = −32/−4 = 8
    $$ \left\{\begin{array} { l } { y _ { 1 } = - 0 , 5 } \\ { x _ { 1 } = 2 · ( - 0 , 5 ) + 3 = 2 } \end{array} \left\{\begin{array}{l} y_1=-0,5 \\ x_1=2 \end{array}\right.\right. $$
    $$ \left\{\begin{array} { l } { y _ { 2 } = 8 } \\ { x _ { 2 } = 2 · 8 + 3 = 1 9 } \end{array} \left\{\begin{array}{l} y_2=8 \\ x_2=19 \end{array}\right.\right. $$
    (2; −0,5), (19; 8)