Question

помогите пожалуйста (только 4-8) даю 50 баллов!​

Answer 1

Ответ:

Применяем метод сложения для решения систем уравнений .

$4)\ \ \left\{\begin{array}{l}y^2+xy=1\\4y-xy=-4\end{array}\right\ \oplus \ \left\{\begin{array}{l}y^2+4y=-3\\4y-xy=-4\end{array}\right\ \ \left\{\begin{array}{l}y^2+4y+3=0\\4y-xy=-4\end{array}\right$

$\left\{\begin{array}{l}y_1=-3\ ;\ y_2=-1\\4y-xy=-4\end{array}\right\ \ \left\{\begin{array}{l}y_1=-3\ ;\ y_2=-1\\-12+3x=-4\ ;\ -4+x=-4\end{array}\right\ \ \left\{\begin{array}{l}y_1=-3\ ,;y_2=-1\\x_1=\frac{8}{3}\ ;\ x_2=0\end{array}\right\\\\\\Otvet:\ (\frac{8}{3}\ ;\ -3\ )\ ,\ (\ 0\, ;-1\, )\ .$

$5)\ \ \left\{\begin{array}{l}x^2+xy=5\\y^2+xy=20\end{array}\right\ \oplus \ \left\{\begin{array}{l}x^2+2xy+y^2=25\\y^2+xy=20\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2-5^2=0\\y^2+xy=20\end{array}\right$

$\left\{\begin{array}{l}(x+y-5)(x+y+5)=0\\y^2+xy=20\end{array}\right\ \ \left\{\begin{array}{l}x_1=5-y\ ;\ x_2=-y-5\\y^2+xy=20\end{array}\right$

$a)\ \ \left\{\begin{array}{l}x_1=5-y\\y^2+y(5-y)=20\end{array}\right\ \ \left\{\begin{array}{l}x_1=5-y\\5y=20\end{array}\right\ \ \left\{\begin{array}{l}x_1=1\\y_1=4\end{array}\right$  

$b)\ \ \left\{\begin{array}{l}x_2=-y-5\\y^2+y(-y-5)=20\end{array}\right\ \ \left\{\begin{array}{l}x_2=-y-5\\-5y=20\end{array}\right\ \ \left\{\begin{array}{l}x_2=-1\\y_2=-4\end{array}\right\\\\\\Otvet:\ \ (\ 1\ ;\ 4\ )\ ,\ (-1\ ,-4\ )\ .$

$6)\ \ \left\{\begin{array}{l}x^2-xy=35\\y^2-xy=14\end{array}\right\ \oplus \ \left\{\begin{array}{l}x^2-2xy+y^2=49\\y^2-xy=14\end{array}\right\ \ \left\{\begin{array}{l}(x-y)^2-7^2=0\\y^2-xy=14\end{array}\right\\\\\\\left\{\begin{array}{l}(x-y-7)(x-y+7)=0\\y^2-xy=14\end{array}\right\ \ \left\{\begin{array}{l}x=y+7\ ;\ x=y-7\\y^2-xy=14\end{array}\right$

$a)\ \ \left\{\begin{array}{l}x=y+7\\y^2-y(y+7)=14\end{array}\right\ \ \left\{\begin{array}{l}x=y+7\\-7y=14\end{array}\right\ \ \left\{\begin{array}{l}x=5\\y=-2\end{array}\right$

$b)\ \ \left\{\begin{array}{l}x=y-7\\y^2-y(y-7)=14\end{array}\right\ \ \left\{\begin{array}{l}x=y-7\\7y=14\end{array}\right\ \ \left\{\begin{array}{l}x=-5\\y=2\end{array}\right\\\\\\Otvet:\ \ (\ 5\ ;-2\ )\ ,\ (\ -5\ ;\ 2\ )\ .$

$7)\ \ \left\{\begin{array}{l}9x^2-5xy=-6\\y^2-xy=6\end{array}\right\ \oplus \ \left\{\begin{array}{l}9x^2-6xy+y^2=0\\y^2-xy=6\end{array}\right\ \ \left\{\begin{array}{l}(3x-y)^2=0\\y^2-xy=6\end{array}\right\\\\\\\left\{\begin{array}{l}3x-y=0\\y^2-xy=6\end{array}\right\ \ \left\{\begin{array}{l}y=3x\\9x^2-3x^2=6\end{array}\right\ \ \left\{\begin{array}{l}y=3x\\6x^2=6\end{array}\right\ \ \left\{\begin{array}{l}y=3x\\x^2=1\end{array}\right$

$\left\{\begin{array}{l}y_1=-3\ ,\ y_2=3\\x_1=-1\ ,\ x_2=1\end{array}\right\ \ \ \ \ Otvet:\ \ (-1\ ;-3\ )\ ,\ (\ 1\ ;\ 3\ )\ .$

$8)\ \ \left\{\begin{array}{l}4x^2-xy=8\\y^2-3xy=-8\end{array}\right\ \oplus \ \left\{\begin{array}{l}4x^2-4xy+y^2=0\\y^2-3xy=-8\end{array}\right\ \ \left\{\begin{array}{l}(2x-y)^2=0\\y^2-3xy=-8\end{array}\right\\\\\\\left\{\begin{array}{l}2x-y=0\\y^2-3xy=-8\end{array}\right\ \ \left\{\begin{array}{l}y=2x\\4x^2-6x^2=-8\end{array}\right\ \ \left\{\begin{array}{l}y=2x\\-2x^2=-8\end{array}\right\ \ \left\{\begin{array}{l}y=2x\\x^2=4\end{array}\right$

$\left\{\begin{array}{l}y_1=-4\ ,\ y_2=4\\x_1=-2\ ,\ x_2=2\end{array}\right\ \ \ \ \ Otvet:\ \ (-2\ ;-4\ )\ ,\ (\ 2\ ;\ 4\ )\ .$