\(HPT\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-2xy=11\\\left(x+y\right)+xy=3+4\sqrt{2}\end{cases}}\)
Đặt x+y=a;xy=b thì hệ trở thành:
\(\hept{\begin{cases}a^2-2b=11\\a+b=3+4\sqrt{2}\end{cases}\Leftrightarrow\hept{\begin{cases}a^2-2b=11\\b=3+4\sqrt{2}-a\end{cases}}}\)
=> \(a^2-2\left(3+4\sqrt{2}-a\right)=11\)
<=>\(a^2-6-8\sqrt{2}+2a-11=0\)
\(\Leftrightarrow a^2+2a-17-8\sqrt{2}=0\)
\(\Leftrightarrow a^2+2a-\left(16+2.4.\sqrt{2}+2-1\right)=0\)
\(\Leftrightarrow\left(a+1\right)^2-\left(4+\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(a+1-4-\sqrt{2}\right)\left(a+1+4+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(a-3-\sqrt{2}\right)\left(a+5+\sqrt{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-3-\sqrt{2}=0\\a+5+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3+\sqrt{2}\\a=-5-\sqrt{2}\end{cases}}}\)
=>\(\orbr{\begin{cases}b=3+4\sqrt{2}-3-\sqrt{2}=3\sqrt{2}\\b=3+4\sqrt{2}+5+\sqrt{2}=8+5\sqrt{2}\end{cases}}\)
- Với \(a=3+\sqrt{2},b=3\sqrt{2}\),ta có: \(x+y=3+\sqrt{2}\Rightarrow y=3+\sqrt{2}-x\) (1)
Thay (1) vào \(xy=3\sqrt{2}=b\Rightarrow x\left(3+\sqrt{2}-x\right)=3\sqrt{2}\)
\(\Leftrightarrow3x+x\sqrt{2}-x^2=3\sqrt{2}\Leftrightarrow x\left(3-x\right)+\sqrt{2}\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\sqrt{2}-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\sqrt{2}\\y=3\end{cases}}\)
- Với \(a=-5-\sqrt{2},b=8+5\sqrt{2}\), ta có: \(x+y=-5-\sqrt{2}\Rightarrow y=-5-\sqrt{2}-x\)(2)
Thay (2) vào \(xy=8+5\sqrt{2}=b\Rightarrow x\left(-5-\sqrt{2}-x\right)=8+5\sqrt{2}\)
\(\Leftrightarrow-x^2-5x-x\sqrt{2}-8-5\sqrt{2}=0\)
\(\Leftrightarrow-x^2+\left(-5-\sqrt{2}\right)x+\left(-8-5\sqrt{2}\right)=0\)(3)
\(\Delta=\left(-5-\sqrt{2}\right)^2-4.\left(-1\right).\left(-8-5\sqrt{2}\right)\)
\(=27+10\sqrt{2}-32-20\sqrt{2}=-5-10\sqrt{2}< 0\)
=>pt (3) vô nghiệm
Vậy \(\left(x;y\right)=\left(3;\sqrt{2}\right)\) hoaojwc \(\left(\sqrt{2};3\right)\)
