Решите систему уравнений:
а) \(\left\{\begin{array}{l}x+3y=-1 \\x^{2}+2xy+y=3\end{array}\right.\)
б) \(\left\{\begin{array}{l}2x-y=1 \\xy-y^{2}+3x=-1\end{array}\right.\)
в) \(\left\{\begin{array}{l}2x+y-11=0, \\2x+5y-y^{2}-6=0\end{array}\right.\)
г) \(\left\{\begin{array}{l}2x^{2}-3y^{2}-5x-2y=26, \\x-y=4\end{array}\right.\)
д) \(\left\{\begin{array}{l}4x^{2}-9y^{2}+x-40y=19 \\2x-3y=5\end{array}\right.\)
e) \(\left\{\begin{array}{l}3x^{2}+y^{2}+8x+13y=5, \\x-y+2=0\end{array}\right.\)
\(\begin{cases}x+3 y=-1 \\x^2+2 x y+y=3\end{cases}\)
\(\begin{cases}x=-1-3 y\\(1+3 y)^2-2 y(1+3 y)+y-3=0\end{cases}\)
\(\begin{cases}x=-1-3 y\\1+6 y+9 y^2-2 y-6 y^2+y-3=0\end{cases} \)
\(3 y^2+5 y-2=0\\ y_{1,2}=\frac{-5 \pm \sqrt{25+24}}{6}\\ y_1=\frac{1}{3} \\ y_2=-2\)
\(\begin{cases}{ y = \frac { 1 } { 3 } } \\{ x = - 2 }\end{cases}\) или \(\begin{cases}y=-2 \\x=5\end{cases}\)
Ответ: \((-2; \frac{1}{3}),(5;-2)\).
\(\begin{cases}2 x-y=1 \\x y-y^2+3 x=-1\end{cases} \)
\(\begin{cases}y=2 x-1 \\x(2 x-1)-(2 x-1)^2+3 x=-1\end{cases}\)
\(\begin{cases}y=2 x-1 \\2 x^2-x-4 x^2+4 x-1+3 x+1=0\end{cases} \)
\(-2 x^2+6 x=0 \\ x(x-3)=0 \\ x_1=0 \\ x_2=3 \)
\(\begin{cases}{ x = 0 } \\{ y = - 1 }\end{cases} \) или \(\begin{cases}x=3 \\y=5\end{cases}\)
Ответ: \((0;-1),(3; 5)\).
\(\begin{cases}2 x+y-11=0 \\2 x+5 y-y^2-6=0\end{cases} \)
\(\begin{cases}y=11-2 x \\2 x+5(11-2 x)-(11-2 x)^2-6=0\end{cases}\)
\(\begin{cases}y=11-2 x \\2 x+55-10 x-121+44 x-4 x^2-6=0\end{cases} \)
\(4 x^2-36 x+72=0 \\ x^2-9 x+18=0 \\ x_{1,2}=\frac{9 \pm \sqrt{81-72}}{2} \\ x_1=6 \\ x_2=3\)
\(\begin{cases}{ x = 6 } \\{ y = - 1 }\end{cases} \) или \(\begin{cases}x=3 \\y=5\end{cases}\)
Ответ: \((6;-1),(3; 5)\).
\(\begin{cases}2 x^2-3 y^2-5 x-2 y=26 \\x-y=4\end{cases} \)
\(\begin{cases}y=x-4 \\2 x^2-3(x-4)^2-5 x-2(x-4)=26\end{cases} \)
\(\begin{cases}y=x-4 \\2 x^2-3 x^2+24 x-48-5 x-2 x+8-26=0\end{cases}\)
\(x^2-17 x+66=0 \\ x_{1,2}=\frac{17 \pm \sqrt{289-264}}{2} \\ x_1=11 \\ x_2=6 \)
\(\begin{cases}x = 1 1 \\y = 7\end{cases} \) или \(\begin{cases}x=6 \\y=2\end{cases}\)
Ответ: \((11; 7),(6; 2)\).
\( \begin{cases}4 x^2-9 y^2+x-40 y=19 \\2 x-3 y=5 \end{cases} \)
\(\begin{cases}x=2,5+1,5 y\\4(2,5+1,5 y)^2-9 y^2+2,5+1,5 y-40 y-19=0\end{cases}\)
\(\begin{cases}x=2,5+1,5 y\\25+30 y+9 y^2-9 y^2-38,5 y-16,5=0\end{cases}\)
\(\begin{cases}x=2,5+1,5 y \\8,5-8,5 y=0\end{cases}\)
\(\begin{cases}y=1 \\x=4\end{cases}\)
Ответ: \((4; 1)\).
\(\begin{cases}3 x^2+y^2+8 x+13 y=5 \\x-y+2=0\end{cases}\)
\(\begin{cases}x= y-2\\ 3(y-2)^2+y^2+8(y-2)+13y-5=0\end{cases}\)
\(\begin{cases}x=y-2 \\3 y^2-12 y+12+y^2+8 y-16+13 y-5=0\end{cases} \)
\(4 y^2+9 y-9=0\\ y_{1,2}=\frac{-9 \pm \sqrt{81+144}}{8} \\ y_1=\frac{3}{4} \\ y_2=-3 \)
\(\begin{cases}y = \frac { 3 } { 4 }\\x = - \frac { 5 } { 4 } \end{cases} \) или \(\begin{cases}y=-3 \\x=-5\end{cases}\)
Ответ: \((-1 \frac{1}{4}; \frac{3}{4}),( -5;-3)\).
Решебник
"Алгебра - Учебник" по предмету Математика за 9 класс.
Aвторы:
Макарычев Ю.Н., Миндюк Н.Г., Нешков К.И., Суворова С.Б.
Задание
Решите систему уравнений: а) \(\left\{\begin{array}{l}x+3y=-1 \\x^{2}+2xy+y=3\end{array}\right.\) б) \(\left\{\begin{array}{l}2x-y=1 \\xy-y^{2}+3x=-1\end{array}\right.\) в) \(\left\{\begin{array}{l}2x+y-11=0, \\2x+5y-y^{2}-6=0\end{array}\right.\) г) \(\left\{\begin{array}{l}2x^{2}-3y^{2}-5x-2y=26, \\x-y=4\end{array}\right.\) д) \(\left\{\begin{array}{l}4x^{2}-9y^{2}+x-40y=19 \\2x-3y=5\end{array}\right.\) e) \(\left\{\begin{array}{l}3x^{2}+y^{2}+8x+13y=5, \\x-y+2=0\end{array}\right.\)