Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 707

а) $$ \left\{\begin{array}{l} 2 x^2+y^2=9 \\ x^2-y^2=3 \end{array}\right. $$
3x² = 12 x² = 12 : 3 x² = 4 x = ±$$\sqrt{4}$$ x = ±2
$$ \left\{\begin{array}{l} x_1=2 \\ y_1=\sqrt{2^2-3}=\sqrt{1}= \pm 1 \end{array}\right. $$
$$ \left\{\begin{array}{l} x_2=-2 \\ y_2=\sqrt{2^2-3}=\sqrt{1}= \pm 1 \end{array}\right. $$
(2; 1), (2; −1), (−2; 1), (−2; −1)
б) $$ \left\{\begin{array} { l } { 2 x ^ { 2 } - x y = 3 3 } \\ { 4 x - y = 1 7 } \end{array} \left\{\begin{array}{l} 2 x^2-x(4 x-17)=33 \\ y=4 x-17 \end{array}\right.\right. $$ $$ \left\{\begin{array}{l} 2 x^2-4 x^2+17 x-33 \\ y=4 x-17 \end{array}=0\left\{\begin{array}{l} -2 x^2+17 x-33=0 \\ y=4 x-17 \end{array}\right.\right. $$
−2х² + 17х − 33 = 0 D = b² − 4ac = 17² − 4 · (−2) · (−33) = 289 − 264 = 25 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-17+\sqrt{25}}{2(-2)} $$ = −17 + 5/−4 = −12/−4 = 3
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-17-\sqrt{25}}{2(-2)} $$ = −17 − 5/−4 = −22/−4 = 5,5
x₁ = 3, x₂ = 5,5
$$ \left\{\begin{array} { l } { x _ { 1 } = 3 } \\ { y _ { 1 } = 4 · 3 - 1 7 = - 5 } \end{array} \left\{\begin{array}{l} x_1=3 \\ y_1=-5 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = 5 , 5 } \\ { y _ { 2 } = 4 · 5 , 5 - 1 7 = 5 } \end{array} \left\{\begin{array}{l} x_2=5,5 \\ y_2=5 \end{array}\right.\right. $$
(3; −5), (5,5; 5)
в) $$ \left\{\begin{array} { l } { 3 x ^ { 2 } - 2 y = 1 } \\ { 2 x ^ { 2 } - y ^ { 2 } = 1 } \end{array} \left\{\begin{array}{l} x^2=\frac{1+2 y}{3} \\ 2 \frac{1+2 y}{3}-y^2=1 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { x ^ { 2 } = \frac { 1 + 2 y } { 3 } } \\ { \frac { 2 + 4 y } { 3 } - y ^ { 2 } = 1 | 3 } \\ { 2 + 4 y - 3 y ^ { 2 } - 3 = 0 } \end{array} \left\{\begin{array}{l} x^2=\frac{1+2 y}{3}=\frac{1+2 y}{3} \\ -3 y^2+4 y-1=0 \end{array}\right.\right. $$
−3у² + 4у − 1 = 0(−1) 3у² − 4у + 1 = 0 D = b² − 4ac = (−4)² − 4 · 3 · 1 = 16 − 12 = 4 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{4+\sqrt{4}}{2 · 3} $$ = 4 + 2/6 = 6/6 = 1
y₂ = $$ =\frac{-b-\sqrt{D}}{2 a}=\frac{4-\sqrt{4}}{2 · 3} $$ = 4 − 2/6 = 2/6 = 1/3
y₁ = 1, y₂ = 1/3
$$ \left\{\begin{array} { l } { y _ { 1 } = 1 } \\ { x ^ { 2 } = \frac { 1 + 2 · 1 } { 3 } = 1 } \end{array} \left\{\begin{array}{l} y_1=1 \\ x_{1,2}= \pm 1 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = \frac { 1 } { 3 } } \\ { x ^ { 2 } = \frac { 1 + 2 \frac { 1 } { 3 } } { 3 } = \frac { 5 } { 9 } } \end{array} \left\{\begin{array}{l} y_2=\frac{1}{3} \\ x_{3,4}= \pm \frac{\sqrt{5}}{3} \end{array}\right.\right. $$
(1; 1), (−1; 1), $$ \left(\frac{\sqrt{5}}{3} ; \frac{1}{3}\right),\left(-\frac{\sqrt{5}}{3} ; \frac{1}{3}\right) $$
г) $$ \left\{\begin{array} { l } { x - y - 4 = 0 } \\ { x ^ { 2 } + y ^ { 2 } = 8 , 5 } \end{array} \left\{\begin{array}{l} x=y+4 \\ (y+4)^2+y^2=8,5 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { x = y + 4 } \\ { y ^ { 2 } + 8 y + 1 6 + y ^ { 2 } - 8 , 5 = 0 } \end{array} \left\{\begin{array}{l} x=y+4 \\ 2 y^2+8 y+7,5=0 \end{array}\right.\right. $$
2у² + 8у + 7,5 = 0 D = b² − 4ac = 8² − 4 · 2 · 7,5 = 64 − 60 = 4 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-8+\sqrt{4}}{2 · 2} $$ = −8 + 2/4 = −6/4 = −3/2 = −1,5
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-8-\sqrt{4}}{2 · 2} $$ = −8 − 2/4 = −10/4 = −5/2 = −2,5
y₁ = −1,5, y₂ = −2,5
$$ \left\{\begin{array}{l} y_1=-1,5 \\ x_1=-1,5+4 \end{array}=2,5\left\{\begin{array}{l} y_1=-1,5 \\ x_1=2,5 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} y_2=-2,5 \\ x_2=-2,5+4 \end{array}=1,5\left\{\begin{array}{l} y_2=-2,5 \\ x_2=1,5 \end{array}\right.\right. $$
(2,5; −1,5), (1,5; −2,5)
д) $$ \left\{\begin{array} { l } { x ^ { 2 } + 4 y = 1 0 } \\ { x - 2 y = - 5 } \end{array} \left\{\begin{array}{l} (2 y-5)^2+4 y=10 \\ x=2 y-5 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { 4 y ^ { 2 } - 2 0 y + 2 5 + 4 y - 1 0 = 0 } \\ { x = 2 y - 5 } \end{array} \left\{\begin{array}{l} 4 y^2-16 y+15=0 \\ x=2 y-5 \end{array}\right.\right. $$
4у² − 16у + 15 = 0 D = b² − 4ac = (−16)² − 4 · 4 · 15 = 256 − 240 = 16 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{16+\sqrt{16}}{2 · 4} $$ = 16 + 4/8 = 20/8 = 5/2 = 2,5
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{16-\sqrt{16}}{2 · 4} $$ = 16 − 4/8 = 12/8 = 3/2 = 1,5
y₁ = 2,5, y₂ = 1,5
$$ \left\{\begin{array} { l } { y _ { 1 } = 2 , 5 } \\ { x _ { 1 } = 2 2 , 5 - 5 = 0 } \end{array} \left\{\begin{array}{l} y_1=2,5 \\ x_1=0 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = 1 , 5 } \\ { x _ { 2 } = 2 \unrhd 1 , 5 - 5 = - 2 } \end{array} \left\{\begin{array}{l} y_2=1,5 \\ x_2=-2 \end{array}\right.\right. $$
(0; 2,5), (−2; 1,5)
е) $$ \left\{\begin{array} { l } { x - 2 y + 1 = 0 } \\ { 5 x y + y ^ { 2 } = 1 6 } \end{array} \left\{\begin{array}{l} x=2 y-1 \\ 5 y(2 y-1)+y^2=16 \end{array}\right.\right. $$ $$ \left\{\begin{array} { l } { x = 2 y - 1 } \\ { 1 0 y ^ { 2 } - 5 y + y ^ { 2 } - 1 6 = 0 } \end{array} \left\{\begin{array}{l} x=2 y-1 \\ 11 y^2-5 y-16=0 \end{array}\right.\right. $$
11у² − 5у − 16 = 0 D = b² − 4ac = (−5)² − 4 · 11 · (−16) = 25 + 704 = 729 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{5+\sqrt{729}}{2 · 11} $$ = 5 + 27/22 = 32/22 = 16/11 = 15/11
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{5-\sqrt{729}}{2 · 11} $$ = 5 − 27/22 = −22/22 = −1
y₁ = 15/11, y₂ = −1
$$ \left\{\begin{array} { l } { y _ { 1 } = 1 \frac { 5 } { 1 1 } } \\ { x _ { 1 } = 2 \frac { 1 6 } { 1 1 } - 1 = 1 \frac { 1 0 } { 1 1 } } \end{array} \left\{\begin{array}{l} y_1=1 \frac{5}{11} \\ x_1=1 \frac{10}{11} \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 1 } \\ { x _ { 2 } = 2 ( - 1 ) - 1 = - 3 } \end{array} \left\{\begin{array}{l} y_2=-1 \\ x_2=-3 \end{array}\right.\right. $$
(110/11; 15/11), (−3; −1)