Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 706

а) $$ \left\{\begin{array} { l } { y - 2 x = 2 } \\ { 5 x ^ { 2 } - y = 1 } \end{array} \left\{\begin{array} { l } { y = 2 + 2 x } \\ { 5 x ^ { 2 } - 2 - 2 x - 1 = 0 } \end{array} \left\{\begin{array}{l} y=2+2 x \\ 5 x^2-2 x-3=0 \end{array}\right.\right.\right. $$
5х² − 2х − 3 = 0 D = b² − 4ac = (−2)² − 4 ⋅ 5 ⋅ (−3) = 4 + 60 = 64 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{2+\sqrt{64}}{2 ⋅ 5} $$ = 2 + 8/10 = 10/10 = 1
x₂ = $$ =\frac{-b-\sqrt{D}}{2 a}=\frac{2-\sqrt{64}}{2 ⋅ 5} $$ = 2 − 8/10 = −6/10 = −0,6
x₁ = 1, x₂ = −0,6
$$ \left\{\begin{array} { l } { x _ { 1 } = 1 } \\ { y _ { 1 } = 2 + 2 ⋅ 1 = 4 } \end{array} \left\{\begin{array}{l} x_1=1 \\ y_1=4 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = - 0 , 6 } \\ { y _ { 2 } = 2 + 2 ( - 0 , 6 ) = 0 , 8 } \end{array} \left\{\begin{array}{l} x_2=-0,6 \\ y_2=0,8 \end{array}\right.\right. $$
(1; 4), (−0,6; 0,8)
б) $$ \left\{\begin{array} { l } { x - 2 y ^ { 2 } = 2 } \\ { 3 x + y = 7 } \end{array} \left\{\begin{array}{l} x=2+2 y^2 \\ 3\left(2+2 y^2\right)+y=7 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { x = 2 + 2 y ^ { 2 } } \\ { 6 + 6 y ^ { 2 } + y - 7 = 0 } \end{array} \left\{\begin{array}{l} x=2+2 y^2 \\ 6 y^2+y-1=0 \end{array}\right.\right. $$
6у² + у − 1 = 0 D = b² − 4ac = 1² − 4 ⋅ 6 ⋅ (−1) = 1 + 24 = 25 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-1+\sqrt{25}}{2 ⋅ 6}= $$ = −1 + 5/12 = 4/12 = 1/3
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-1-\sqrt{25}}{2 ⋅ 6} $$ = −1 − 5/12 = −6/12 = −1/2 = −0,5
y₁ = 1/3, y₂ = −0,5
$$ \left\{\begin{array} { l } { y _ { 1 } = \frac { 1 } { 3 } } \\ { x _ { 1 } = 2 + 2 ( \frac { 1 } { 3 } ) ^ { 2 } = 2 \frac { 2 } { 9 } } \end{array} \left\{\begin{array}{l} y_1=\frac{1}{3} \\ x_1=2 \frac{2}{9} \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 0 , 5 } \\ { x _ { 2 } = 2 + 2 \square ( - 0 , 5 ) ^ { 2 } = 2 , 5 } \end{array} \left\{\begin{array}{l} y_2=-0,5 \\ x_2=2,5 \end{array}\right.\right. $$
(22/9; 1/3), (2,5; −0,5)
в) $$ \left\{\begin{array} { l } { 3 x ^ { 2 } - 2 y = 1 } \\ { 2 x ^ { 2 } - y ^ { 2 } = 1 } \end{array} \left\{\begin{array}{l} 2 y=3 x^2-1 \\ 2 x^2-y^2=1 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { y = \frac { 3 x ^ { 2 } - 1 } { 2 } } \\ { 2 x ^ { 2 } - y ^ { 2 } = 1 } \end{array} \left\{\begin{array}{l} y=1,5 x^2-0,5 \\ 2 x^2-\left(1,5 x^2-0,5\right)^2=1 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} y=1,5 x^2-0,5 \\ 2 x^2-2,25 x^2+1,5 x-0,25-1=0 \end{array}\right. $$ $$ \left\{\begin{array}{l} y=1,5 x^2-0,5 \\ -0,25 x^2+1,5 x-1,25=0 \end{array}\right. $$
−0,25х² + 1,5х − 1,25 = 0 : (−0,25) х² − 6х + 5 = 0 D = b² − 4ac = (−6)² − 4 ⋅ 1 ⋅ 5 = 36 − 20 = 16 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{6+\sqrt{16}}{2 ⋅ 1} $$ = 6 + 4/2 = 10/2 = 5
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{6-\sqrt{16}}{2 ⋅ 1} $$ = 6 − 4/2 = 2/2 = 1
x₁ = 5, x₂ = 1
$$ \left\{\begin{array} { l } { x _ { 1 } = 5 } \\ { y _ { 1 } = 1 , 5 5 ^ { 2 } - 0 , 5 = 3 7 } \end{array} \left\{\begin{array}{l} x_1=5 \\ y_1=37 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = 1 } \\ { y _ { 2 } = 1 , 5 ⋅ ^ { 2 } - 0 , 5 = 1 } \end{array} \left\{\begin{array}{l} x_2=1 \\ y_2=1 \end{array}\right.\right. $$
(5; 37), (1; 1)
г) $$ \left\{\begin{array} { l } { 3 x ^ { 2 } + 2 y ^ { 2 } = 1 1 } \\ { x + 2 y = 3 } \end{array} \left\{\begin{array}{l} 3(3-2 y)^2+2 y^2=11 \\ x=3-2 y \end{array}\right.\right. $$ $$ \left\{\begin{array}{l} 3\left(9-12 y+4 y^2\right)+2 y^2=11 \\ x=3-2 y \end{array}\right. $$
$$ \left\{\begin{array} { l } { 2 7 - 3 6 y + 1 2 y ^ { 2 } + 2 y ^ { 2 } - 1 1 = 0 } \\ { x = 3 - 2 y } \end{array} \left\{\begin{array}{l} 14 y^2-36 y+16=0 \\ x=3-2 y \end{array}\right.\right. $$
14у² − 36у + 16 = 0 : 2 7у² − 18у + 8 = 0 D = b² − 4ac = (−18)² − 4 ⋅ 7 ⋅ 8 = 324 − 224 = 100 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{18+\sqrt{100}}{2 ⋅ 7} $$ = 18 + 10/14 = 28/14 = 2
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{18-\sqrt{100}}{2 ⋅ 7} $$ = 18 − 10/14 = 8/14 = 4/7
y₁ = 2, y₂ = 4/7
$$ \left\{\begin{array} { l } { y _ { 1 } = 2 } \\ { x _ { 1 } = 3 - 2 2 = - 1 } \end{array} \left\{\begin{array}{l} y_1=2 \\ x_1=-1 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = \frac { 4 } { 7 } } \\ { x _ { 2 } = 3 - 2 \frac { 4 } { 7 } = 2 \frac { 6 } { 7 } } \end{array} \left\{\begin{array}{l} y_2=\frac{4}{7} \\ x_2=2 \frac{6}{7} \end{array}\right.\right. $$
(−1; 2), (26/7; 4/7)
д) $$ \left\{\begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 0 0 } \\ { 3 x = 4 y } \end{array} \left\{\begin{array}{l} \left(\frac{4 y}{3}\right)^2+y^2=100 \\ x=\frac{4 y}{3} \end{array}\right.\right. $$ $$ \left\{\begin{array}{l} \frac{16 y^2}{9}+y^2=100 \left\lvert\, 9\left\{\begin{array}{l} 16 y^2+9 y^2-900=0 \\ x=\frac{4 y}{3} \end{array}\right.\right. \\ x=\frac{4 y}{3} \end{array}\right. $$
$$ \left\{\begin{array} { l } { 2 5 y ^ { 2 } - 9 0 0 = 0 } \\ { x = \frac { 4 y } { 3 } } \end{array} \left\{\begin{array}{l} 25 y^2=900 \\ x=\frac{4 y}{3} \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { y ^ { 2 } = \frac { 9 0 0 } { 2 5 } } \\ { x = \frac { 4 y } { 3 } } \end{array} \left\{\begin{array} { l } { y = \pm \sqrt { 3 6 } } \\ { x = \frac { 4 y } { 3 } } \end{array} \left\{\begin{array}{l} y= \pm 6 \\ x=\frac{4 y}{3} \end{array}\right.\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 1 } = 6 } \\ { x _ { 1 } = \frac { 4 ⋅ 6 } { 3 } = 4 ⋅ 2 = 8 } \end{array} \left\{\begin{array}{l} y_1=6 \\ x_1=8 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 1 } = - 6 } \\ { x _ { 1 } = \frac { 4 ⋅ ( - 6 ) } { 3 } = 4 ⋅ ( - 2 ) = - 8 } \end{array} \left\{\begin{array}{l} y_1=-6 \\ x_1=-8 \end{array}\right.\right. $$
(8; 6), (−8; −6)
е) $$ \left\{\begin{array} { l } { 2 x ^ { 2 } - y ^ { 2 } = 3 2 } \\ { 2 x - y = 8 } \end{array} \left\{\begin{array}{l} 2 x^2-(2 x-8)^2=32 \\ y=2 x-8 \end{array}\right.\right. $$ $$ \left\{\begin{array}{l} 2 x^2-4 x^2+32 x-64-32 \\ y=2 x-8 \end{array}=0\left\{\begin{array}{l} -2 x^2+32 x-96=0 \\ y=2 x-8 \end{array}\right.\right. $$
−2х² + 32х − 96 = 0 : (−2) х² − 16х + 48 = 0 D = b² − 4ac = (−16)² − 4 ⋅ 1 ⋅ 48 = 256 − 192 = 64 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{16+\sqrt{64}}{2 ⋅ 1} $$ = 16 + 8/2 = 24/2 = 12
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{16-\sqrt{64}}{2 ⋅ 1} $$ = 16 − 8/2 = 8/2 = 4
x₁ = 12, x₂ = 4
$$ \left\{\begin{array} { l } { x _ { 1 } = 1 2 } \\ { y _ { 1 } = 2 ⋅ 12 - 8 = 1 6 } \end{array} \left\{\begin{array}{l} x_1=12 \\ y_1=16 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = 4 } \\ { y _ { 2 } = 2 \square 4 - 8 = 0 } \end{array} \left\{\begin{array}{l} x_2=4 \\ y_2=0 \end{array}\right.\right. $$
(12; 16), (4; 0)