Question

СРОЧНО ПОМОГИТЕ НОМЕР 129

Answer 1

Ответ:

$1)\ \ \left\{\begin{array}{l}x=5-y\\y^2+4xy=33\end{array}\right\ \ \left\{\begin{array}{l}x=5-y\\y^2+4y(5-y)=33\end{array}\right\ \ \left\{\begin{array}{l}x=5-y\\-3y^2+20y=33\end{array}\right$

$\left\{\begin{array}{l}x=5-y\\3y^2-20y+33=0\end{array}\right\ \ \left\{\begin{array}{l}x=5-y\\(y-3)(3y-11)=0\end{array}\right\ \ \left\{\begin{array}{l}x_1=2\ ,\ x_2=\dfrac{4}{3}\\y_1=3\ ,\ y_2=\dfrac{11}{3}\end{array}\right\\\\\\Otvet:\ \ \Big(\ 2\ ;\ 3\ \Big)\ ,\ \Big(\ \dfrac{4}{3} \ ;\ \dfrac{11}{3}\ \Big)\ .$

$2)\ \ \left\{\begin{array}{l}x+y=8\\xy=-20\end{array}\right\ \ \left\{\begin{array}{l}y=8-x\\x(8-x)=-20\end{array}\right\ \ \left\{\begin{array}{l}y=8-x\\x^2-8x-20=0\end{array}\right$

$\left\{\begin{array}{l}y_1=10\ ,\ \ y_2=-2\\x_1=-2\ ,\ \ x_2=10\end{array}\right\ \ \left\{\begin{array}{l}\end{array}\right\ \ \ \ Otvet:\ \ (-2;10)\ ,\ (10;-2)\ .$

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

$\left\{\begin{array}{l}y=7x+3\\6x^2+24x+18=0\end{array}\right\ \ \left\{\begin{array}{l}y=7x+3\\x^2+4x+3=0\end{array}\right\ \ \left\{\begin{array}{l}y_1=-18\ ,\ y_2=-4\\x_1=-3\ ,\ x_2=-1\end{array}\right\\\\\\Otvet:\ \ (-3;-10)\ ,\ \ (-1;-4)\ .$

$4)\ \ \left\{\begin{array}{l}y^2-xy+x=2\\5y+x=12\end{array}\right\ \ \left\{\begin{array}{l}y^2-(12-5y)\cdot y+12-5y=2\\x=12-5y\end{array}\right$

$\left\{\begin{array}{l}y^2-12y+5y^2+12-5y-2=0\\x=12-5y\end{array}\right\ \ \left\{\begin{array}{l}6y^2-17y+10=0\\x=12-5y\end{array}\right$

$\left\{\begin{array}{l}y_1=\dfrac{5}{6} \ ,\ y_2=2\\\ x_1=\dfrac{47}{6}\ ,\ x_2=2\end{array}\right\ \ \ \ \ Otvet:\ \ \Big(\ \dfrac{47}{6}\ ;\ \dfrac{5}{6}\ \Big)\ ,\ (\ 2\ ;\ 2\ )\ .$