Question

[Pham thi thu Phuong]

Giải hệ phương trình

\(\hept{\begin{cases}xy+x+y=2+3\sqrt{2}\\x^2+y^2=6\end{cases}}\)

Answer 1

[Nyatmax]

\(\Leftrightarrow\hept{\begin{cases}2xy+2x+2y=4+6\sqrt{2}\\x^2+y^2=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2+2\left(x+y\right)-10-6\sqrt{2}=0\left(1\right)\\x^2+y^2=6\left(2\right)\end{cases}}\)

Dat \(x+y=t\left(t\in R\right)\)

PT(1) tro thanh 

\(t^2+2t-10-6\sqrt{2}=0\)

Ta lai co:

\(\Delta^`=1^2-1.\left(-10-6\sqrt{2}\right)=\left(3+\sqrt{2}\right)^2>0\)

\(\Rightarrow t_1=2+\sqrt{2};t_2=-4-\sqrt{2}\)

Voi \(x+y=2+\sqrt{2}\Rightarrow y=2+\sqrt{2}-x\)

Thay vao PT(2) ta duoc:

\(x^2+\left(2+\sqrt{2}-x\right)^2=6\)

\(\Leftrightarrow2x^2-\left(2\sqrt{2}+4\right)x+4\sqrt{2}=0\)

Ta lai co:

\(\Delta^`=\left[-\left(\sqrt{2}+2\right)\right]^2-2.4\sqrt{2}=\left(2-\sqrt{2}\right)^2>0\)

\(\Rightarrow x_1=2;x_2=\sqrt{2}\)

\(\Rightarrow y_1=\sqrt{2};y_2=2\)

Voi \(x+y=-4-\sqrt{2}\Rightarrow y=-4-\sqrt{2}-x\)

Thay vao PT(2) ta duoc:

\(x^2+\left(-4-\sqrt{2}-x\right)^2=6\)

\(\Leftrightarrow2x^2+\left(2\sqrt{2}+8\right)x+12+8\sqrt{2}=0\)

Ta lai co:

\(\Delta^`=\left(\sqrt{2}+4\right)^2-2.\left(12+8\sqrt{2}\right)=-\left(6+8\sqrt{2}\right)< 0\)

Suy ra: \(x+y=-4-\sqrt{2}\left(l\right)\)

Vay nghiem cua HPT la \(\left(2;\sqrt{2}\right),\left(\sqrt{2};2\right)\)