Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 704

а) $$ \left\{\begin{array} { l } { x - y = 3 } \\ { x y = - 2 } \end{array} \left\{\begin{array} { l } { x = y + 3 } \\ { ( y + 3 ) y = - 2 } \end{array} \left\{\begin{array}{l} x=3+y \\ y^2+3 y+2=0 \end{array}\right.\right.\right. $$
у² + 3у + 2 = 0 D = b² − 4ac = 3² − 4 ⋅ 1 ⋅ 2 = 9 − 8 = 1 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-3+\sqrt{1}}{2 ⋅ 1} $$ = −3 + 1/2 = −2/2 = −1 y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-3-\sqrt{1}}{2 ⋅ 1}= $$ = −3 − 1/2 = −4/2 = −2
y₁ = −1, y₂ = −2
$$ \left\{\begin{array} { l } { y _ { 1 } = - 1 } \\ { x _ { 1 } = 3 - 1 = 2 } \end{array} \left\{\begin{array}{l} y_1=-1 \\ x_1=2 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} y_2=-2 \\ x_2=3-2 \end{array}=1\left\{\begin{array}{l} y_2=-2 \\ x_2=1 \end{array}\right.\right. $$
(2; −1), (1; −2)
б) $$ \left\{\begin{array} { l } { x + y = 2 , 5 } \\ { x y = 1 , 5 } \end{array} \left\{\begin{array} { l } { x = 2 , 5 - y } \\ { ( 2 , 5 - y ) y = 1 , 5 } \end{array} \left\{\begin{array}{l} x=2,5-y \\ -y^2+2,5 y-1,5=0 \end{array}\right.\right.\right. $$
−у² + 2,5у − 1,5 = 0 (−1) у² − 2,5у + 1,5 = 0 D = b² − 4ac = (−2,5)² − 4 ⋅ 1 ⋅ 1,5 = 6,25 − 6 = 0,25 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{2,5+\sqrt{0,25}}{2 ⋅ 1} $$ = 2,5 + 0,5/2 = 3/2 = 1,5 y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{2,5-\sqrt{0,25}}{2 ⋅ 1} $$ = 2,5 − 0,5/2 = 2/2 = 1
y₁ = 1,5, y₂ = 1
$$ \left\{\begin{array}{l} y_1=1,5 \\ x_1=2,5-1,5 \end{array}=1\left\{\begin{array}{l} y_1=1,5 \\ x_1=1 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = 1 } \\ { x _ { 2 } = 2 , 5 - 1 = 1 , 5 } \end{array} \left\{\begin{array}{l} y_2=1 \\ x_2=1,5 \end{array}\right.\right. $$
(1; 1,5), (1,5; 1)
в) $$ \left\{\begin{array} { l } { x + y = - 1 } \\ { x ^ { 2 } + y ^ { 2 } = 1 } \end{array} \left\{\begin{array}{l} x=-y-1 \\ (-y-1)^2+y^2=1 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { x = - 1 - y } \\ { y ^ { 2 } + 2 y + 1 + y ^ { 2 } - 1 = 0 } \end{array} \left\{\begin{array}{l} x=-1-y \\ 2 y^2+2 y=0 \end{array}\right.\right. $$
2у² + 2у = 0 2у(у + 1) = 0 у = 0, у + 1 = 0, у = −1
$$ \left\{\begin{array} { l } { y _ { 1 } = 0 } \\ { x _ { 1 } = - 1 - 0 = - 1 } \end{array} \left\{\begin{array}{l} y_1=0 \\ x_1=-1 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} y_2=-1 \\ x_2=-1+1 \end{array}=0\left\{\begin{array}{l} y_2=-1 \\ x_2=0 \end{array}\right.\right. $$
(−1; 0), (0; −1)
г) $$ \left\{\begin{array}{l} x-y=2 \\ x^2-y^2=17 \end{array}\left\{\begin{array}{l} x=2+y \\ (2+y)^2-y^2=17 \end{array}\right\}\right. $$ $$ \left\{\begin{array} { l } { x = 2 + y } \\ { 4 + 4 y + y ^ { 2 } - y ^ { 2 } - 1 7 = 0 } \end{array} \left\{\begin{array}{l} x=2+y \\ 4 y-13=0 \end{array}\right.\right. $$
4у − 13 = 0 4у = 13 у = 13 : 4 у = 3,25
$$ \left\{\begin{array} { l } { y = 3 , 2 5 } \\ { x = 2 + 3 , 2 5 } \end{array} \left\{\begin{array}{l} y=3,25 \\ x=5,25 \end{array}\right.\right. $$
(5,25; 3,25)