Question

способом алгебраического сложения решите систему уравнений ​

Answer 1

Ответ:

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

$\left\{\begin{array}{l}y^2=2x-x^2\\(x-y)^2=2x-x^2-1\end{array}\right\ \ \left\{\begin{array}{l}y^2=2x-x^2\\(x-y)^2=-(x^2-2x+1)\end{array}\right\ \ \left\{\begin{array}{l}y^2=2x-x^2\\(x-y)^2=-(x-1)^2\end{array}\right$

$\left\{\begin{array}{l}y^2=2x-x^2\\(x-y)^2+(x-1)^2=0\end{array}\right\ \ \left\{\begin{array}{l}y^2=2x-x^2\\x-y=0\ ;\ x-1=0\end{array}\right\ \ \left\{\begin{array}{l}y^2=2x-x^2\\y=x\ ;\ x=1\end{array}\right$

$a)\ \left\{\begin{array}{l}y^2=2x-x^2\\y=x\end{array}\right\ \ \left\{\begin{array}{l}x^2=2x-x^2\\y=x\end{array}\right\ \ \left\{\begin{array}{l}2x^2-2x=0\\y=x\end{array}\right\ \ \left\{\begin{array}{l}2x(x-1)=0\\y=x\end{array}\right\\\\\\\left\{\begin{array}{l}x_1=0\ ,\ x_2=1\\y_1=0\ ,\ y_2=1\end{array}\right\ \ \ \ \ (0;0)\ ,\ (1;1)$

$b)\ \ \left\{\begin{array}{l}y^2=2x-x^2\\x=1\end{array}\right\ \ \left\{\begin{array}{l}y^2=2-1\\x=1\end{array}\right\ \ \left\{\begin{array}{l}y^2=1\\x=1\end{array}\right\ \ \left\{\begin{array}{l}y_1=-1\ ,\ y_2=1\\x=1\end{array}\right\\\\\\(1;-1)\ ,\ \ (1;1)$

$Proverka:\ (1;1):\ \left\{\begin{array}{l}1+1=2\\1-2+1=0\end{array}\right\ \ \left\{\begin{array}{l}2=2\\0=0\end{array}\right\ \ \ verno\\\\\\(0;0):\ \left\{\begin{array}{l}0+0=0\\0-0+1=0\end{array}\right\ \ \ \ neverno$

$(1;-1):\ \left\{\begin{array}{l}1+1=2\\1+2+1=0\end{array}\right\ \ \ \ neverno\\\\\\Otvet:\ \ (\ 1\ ;\ 1\ )\ .$