Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 703

а) $$ \left\{\begin{array} { l } { x = 3 - y } \\ { y ^ { 2 } - x = 3 9 } \end{array} \left\{\begin{array}{l} x=3-y \\ y^2-(3-y)=39 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { x = 3 - y } \\ { y ^ { 2 } - 3 + y - 3 9 = 0 } \end{array} \left\{\begin{array}{l} x=3-y \\ y^2+y-42=0 \end{array}\right.\right. $$
у² + у − 42 = 0 D = b² − 4ac = 1² − 4 ⋅ 1 ⋅ (−42) = 1 + 168 = 169 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-1+\sqrt{169}}{2 ⋅ 1} $$ = −1 + 13/2 = 12/2 = 6 y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-1-\sqrt{169}}{2 ⋅ 1} $$ = −1 − 13/2 = −14/2 = −7
y₁ = 6, y₂ = −7
$$ \left\{\begin{array} { l } { y _ { 1 } = 6 } \\ { x _ { 1 } = 3 - 6 = - 3 } \end{array} \left\{\begin{array}{l} y_1=6 \\ x_1=-3 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 7 } \\ { x _ { 2 } = 3 + 7 = 1 0 } \end{array} \left\{\begin{array}{l} y_2=-7 \\ x_2=10 \end{array}\right.\right. $$
(−3; 6), (10; −7)
б) $$ \left\{\begin{array} { l } { y = 1 + x } \\ { x + y ^ { 2 } = - 1 } \end{array} \left\{\begin{array}{l} x=y-1 \\ y-1+y^2+1 \end{array}=0\left\{\begin{array}{l} x=y-1 \\ y+y^2=0 \end{array}\right.\right.\right. $$
у + у² = 0 у(1 + у) = 0 у = 0, 1 + у = 0, у = −1
$$ \left\{\begin{array} { l } { y _ { 1 } = 0 } \\ { x _ { 1 } = 0 - 1 = - 1 } \end{array} \left\{\begin{array}{l} y_1=0 \\ x_1=-1 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 1 } \\ { x _ { 2 } = - 1 - 1 = - 2 } \end{array} \left\{\begin{array}{l} y_2=-1 \\ x_2=-2 \end{array}\right.\right. $$
(−1; 0), (−2; −1)
в) $$ \left\{\begin{array} { l } { x ^ { 2 } + y = 1 4 } \\ { y - x = 8 } \end{array} \left\{\begin{array}{l} x^2+8+x-14 \\ y=8+x \end{array}=0\left\{\begin{array}{l} x^2+x-6=0 \\ y=8+x \end{array}\right.\right.\right. $$
х² + х − 6 = 0 D = b² − 4ac = 1² − 4 ⋅ 1 ⋅ (−6) = 1 + 24 = 25 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-1+\sqrt{25}}{2 ⋅ 1} $$ = −1 + 5/2 = 4/2 = 2 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-1-\sqrt{25}}{2 ⋅ 1} $$ = −1 − 5/2 = −6/2 = −3
x₁ = 2, x₂ = −3
$$ \left\{\begin{array} { l } { x _ { 1 } = 2 } \\ { y _ { 1 } = 8 + 2 = 1 0 } \end{array} \left\{\begin{array}{l} x_1=2 \\ y_1=10 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = - 3 } \\ { y _ { 2 } = 8 - 3 = 5 } \end{array} \left\{\begin{array}{l} x_2=-3 \\ y_2=5 \end{array}\right.\right. $$
(2; 10), (−3; 5)
г) $$ \left\{\begin{array} { l } { x + y = 4 } \\ { y + x y = 6 } \end{array} \left\{\begin{array}{l} y=4-x \\ 4-x+x(4-x)=6 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { y = 4 - x } \\ { 4 - x + 4 x - x ^ { 2 } - 6 = 0 } \end{array} \left\{\begin{array}{l} y=4-x \\ -x^2+3 x-2=0 \end{array}\right.\right. $$
−х² + 3х − 2 = 0 (−1) х² − 3х + 2 = 0 D = b² − 4ac = (−3)² − 4 ⋅ 1 ⋅ 2 = 9 − 8 = 1 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{3+\sqrt{1}}{2 ⋅ 1} $$ = 3 + 1/2 = 4/2 = 2 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{3-\sqrt{1}}{2 ⋅ 1} $$ = 3 − 1/2 = 2/2 = 1
x₁ = 2, x₂ = 1
$$ \left\{\begin{array} { l } { x _ { 1 } = 2 } \\ { y _ { 1 } = 4 - 2 = 2 } \end{array} \left\{\begin{array}{l} x_1=2 \\ y_1=2 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} x_2=1 \\ y_2=4-1 \end{array}=3\left\{\begin{array}{l} x_2=1 \\ y_2=3 \end{array}\right.\right. $$
(2; 2), (1; 3)