Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 711

а) $$ \left\{\begin{array} { l } { x - y = 5 } \\ { \frac { 1 } { x } + \frac { 1 } { y } = \frac { 1 } { 6 } } \end{array} \left\{\begin{array}{l} x=5+y \\ \frac{1}{5+y}+\frac{1}{y}=\frac{1}{6} \end{array} 6 y(5+y) . ~ 0.5 y(5+y)\right.\right. $$ $$ \left\{\begin{array}{l} x=5+y \\ 6 y+6(5+y)=y(5+y) \end{array}\right. $$
$$ \left\{\begin{array} { l } { x = 5 + y } \\ { 6 y + 3 0 + 6 y - 5 y - y ^ { 2 } = 0 } \end{array} \left\{\begin{array}{l} x=5+y \\ y^2-7 y-30=0 \end{array}\right.\right. $$
у² − 7у − 30 = 0 D = b² − 4ac = (−7)² − 4 · 1 · (−30) = 49 + 120 = 169 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{7+\sqrt{169}}{2 · 1} $$ = 7 + 13/2 = 20/2 = 10
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{7-\sqrt{169}}{2 · 1} $$ = 7 − 13/2 = −6/2 = −3
y₁ = 10, y₂ = −3
$$ \left\{\begin{array} { l } { y _ { 1 } = 1 0 } \\ { x _ { 1 } = 5 + 1 0 = 1 5 } \end{array} \left\{\begin{array}{l} y_1=10 \\ x_1=15 \end{array}\right.\right. $$
$$ \left\{\begin{array}{l} y_2=-3 \\ x_2=5-3 \end{array}=2\left\{\begin{array}{l} y_2=-3 \\ x_2=2 \end{array}\right.\right. $$
(15; 10), (2; −3)
б) $$ \left\{\begin{array} { l } { x + y = 6 } \\ { \frac { 1 } { x } - \frac { 1 } { y } = \frac { 1 } { 4 } } \end{array} \left\{\begin{array} { l } { x = 6 - y } \\ { \frac { 1 } { 6 - y } - \frac { 1 } { y } = \frac { 1 } { 4 } · 4 y ( 6 - y ) } \end{array} \left\{\begin{array}{l} x=6-y \\ 4 y-4(6-y)=y(6-y) \end{array}\right.\right.\right. $$
$$ \left\{\begin{array} { l } { x = 6 - y } \\ { 4 y - 2 4 + 4 y - 6 y + y ^ { 2 } = 0 } \end{array} \left\{\begin{array}{l} x=6-y \\ y^2+2 y-24=0 \end{array}\right.\right. $$
у² + 2у − 24 = 0 D = b² − 4ac = 2² − 4 · 1 · (−24) = 4 + 96 = 100 > 0, имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-2+\sqrt{100}}{2 · 1} $$ = −2 + 10/2 = 8/2 = 4
y₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-2-\sqrt{100}}{2 · 1} $$ = −2 − 10/2 = −12/2 = −6
y₁ = 4, y₂ = −6
$$ \left\{\begin{array} { l } { y _ { 1 } = 4 } \\ { x _ { 1 } = 6 - 4 = 2 } \end{array} \left\{\begin{array}{l} y_1=4 \\ x_1=2 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 6 } \\ { x _ { 2 } = 6 + 6 = 1 2 } \end{array} \left\{\begin{array}{l} y_2=-6 \\ x_2=12 \end{array}\right.\right. $$
(2; 4), (12; −6)
в) $$ \left\{\begin{array} { l } { 3 x + y = 1 } \\ { \frac { 1 } { x } + \frac { 1 } { y } = - 2 , 5 } \end{array} \left\{\begin{array} { l } { y = 1 - 3 x } \\ { \frac { 1 } { x } + \frac { 1 } { 1 - 3 x } = - 2 , 5 · x ( 1 - 3 x ) } \end{array} \left\{\begin{array}{l} y=1-3 x \\ 1-3 x+x=-2,5 x+7,5 x^2 \end{array}\right.\right.\right. $$
$$ \left\{\begin{array} { l } { y = 1 - 3 x } \\ { 1 - 3 x + x + 2 , 5 x - 7 , 5 x ^ { 2 } = 0 · 2} \end{array} \left\{\begin{array}{l} y=1-3 x \\ -7,5 x^2+0,5 x+1 \end{array}=0\left\lceil( 2 ) \left\{\begin{array}{l} y=1-3 x \\ 15 x^2-x-2=0 \end{array}\right.\right.\right.\right. $$
15х² − х − 2 = 0 D = b² − 4ac = (−1)² − 4 · 15 · (−2) = 1 + 120 = 121 > 0, имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{1+\sqrt{121}}{2 · 15} $$ = 1 + 11/30 = 12/30 = 2/5 = 0,4
x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{1-\sqrt{121}}{2 · 15} $$ = 1 − 11/30 = −10/30 = −1/3
x₁ = 0,4, x₂ = −1/3
$$ \left\{\begin{array} { l } { x _ { 1 } = 0 , 4 } \\ { y _ { 1 } = 1 - 3 \bullet 0 , 4 = - 0 , 2 } \end{array} \left\{\begin{array}{l} x_1=0,4 \\ y_1=-0,2 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { x _ { 2 } = - \frac { 1 } { 3 } } \\ { y _ { 2 } = 1 + 3 \frac { 1 } { 3 } = 2 } \end{array} \left\{\begin{array}{l} x_2=-\frac{1}{3} \\ y_2=2 \end{array}\right.\right. $$
(0,4; 0,2), (−1/3; 2)
г) $$ \left\{\begin{array}{l} \frac{1}{y}-\frac{1}{x}=\frac{1}{3}\left\{\begin{array} { l } { \frac { 1 } { y } - \frac { 1 } { 2 + 2 y } = \frac { 1 } { 3 } · 3 y ( 2 + 2 y ) } \\ { x - 2 y = 2 } \end{array} \left\{\begin{array}{l} 3(2+2 y)-3 y=y(2+2 y) \\ x=2+2 y \end{array}\right.\right. \end{array}\right. $$
$$ \left\{\begin{array} { l } { 6 + 6 y - 3 y - 2 y - 2 y ^ { 2 } = 0 } \\ { x = 2 + 2 y } \end{array} \left\{\begin{array}{l} -2 y^2+y+6=0 \\ x=2+2 y \end{array}\right.\right. $$
−2у² + у + 6 = 0 : (−1) 2у² − у - 6 = 0 D = b² − 4ac = (−1)² − 4 · 2 · (−6) = 1 + 48 = 49 > 0, имеет 2 корня
y₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{1+\sqrt{49}}{2 · 2} $$ = 1 + 7/4 = 8/4 = 2
y₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{1-\sqrt{49}}{2 · 2} $$ = 1 − 7/4 = −6/4 = −3/4 = −1,5
y₁ = 2, y₂ = −1,5
$$ \left\{\begin{array} { l } { y _ { 1 } = 2 } \\ { x _ { 1 } = 2 + 2 · 2 = 6 } \end{array} \left\{\begin{array}{l} y_1=2 \\ x_1=6 \end{array}\right.\right. $$
$$ \left\{\begin{array} { l } { y _ { 2 } = - 1 , 5 } \\ { x _ { 2 } = 2 + 2 · ( - 1 , 5 ) = - 1 } \end{array} \left\{\begin{array}{l} y_2=-1,5 \\ x_2=-1 \end{array}\right.\right. $$
(6: 2), (−1; −1,5)