Question

мне нужно полное решение и ответ

Answer 1

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

Введём замену:   $u=x+y\ ,\ v=xy$

$\left\{\begin{array}{l}u+4v=6\\v\cdot u=2\end{array}\right\ \ \left\{\begin{array}{l}u=-4v+6\\v\cdot (6-4v)=2\end{array}\right\ \ \left\{\begin{array}{l}u=-4v+6\\4v^2-6v+2=0\ |:2\end{array}\right\ \ \left\{\begin{array}{l}u=-4v+6\\2v^2-3v+1=0\end{array}\right$

Решим квадратное уравнение   $2v^2-3v+1=0\ \,\ \ D=9-4\cdot 2=1\ \ ,\ \ v_1=\dfrac{3-1}{4}=\dfrac{1}{2}\ ,\ \ \ v_2=1$

$\left\{\begin{array}{l}u=-4v+6\\v_1=\dfrac{1}{2}\ ,\ v_2=1\end{array}\right\ \ \left\{\begin{array}{l}u_1=4\ ,\ u_2=2\\v_1=\dfrac{1}{2}\ ,\ v_2=1\end{array}\right\ \ a)\ \ \left\{\begin{array}{l}x+y=4\\xy=\dfrac{1}{2}\end{array}\right\ \ \left\{\begin{array}{l}y=4-x\\x(4-x)-\dfrac{1}{2}=0\end{array}\right\ \ \left\{\begin{array}{l}y=4-x\\x^2-4x+\dfrac{1}{2}=0\end{array}\right\ \ \left\{\begin{array}{l}y=4-x\\2x^2-8x+1=0\end{array}\right$

$\left\{\begin{array}{l}y=4-x\\x_1=\dfrac{4-\sqrt{14}}{2}\ ,\ x_2=\dfrac{4+\sqrt{14}}{2}\end{array}\right\ \ \left\{\begin{array}{l}y_1=2+\dfrac{\sqrt{14}}{2}\ ,\ y_2=2-\dfrac{\sqrt{14}}{2}\\x_1=2-\dfrac{\sqrt{14}}{2}\ ,\ x_2=2+\dfrac{\sqrt{14}}{2}\end{array}\right$

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

$\left\{\begin{array}{l}y=1\\x=1\end{array}\rightOtvet:\ \Big(2-\dfrac{\sqrt{14}}{2}\ ;\ 2+\dfrac{\sqrt{14}}{2}\Big)\ ,\ \Big(2+\dfrac{\sqrt{14}}{2}\ ;\ 2-\dfrac{\sqrt{14}}{2}\Big)\ ,\ \Big(\ 1\ ;\ 1\ \Big)\ .$