\(y=\sqrt{x^2+2x+6}-1\) thay vào pt dưới:
\(x^2+\left(\sqrt{x^2+2x+6}-1\right)^2+x\left(\sqrt{x^2+2x+6}-1\right)=7\)
\(\Leftrightarrow2x^2+x-2\sqrt{x^2+2x+6}+x\sqrt{x^2+2x+6}=0\)
\(\Leftrightarrow2x^2+4x-6+\left(x-2\right)\sqrt{x^2+2x+6}-3\left(x-2\right)=0\)
\(\Leftrightarrow2\left(x^2+2x-3\right)+\left(x-2\right)\left(\sqrt{x^2+2x+6}-3\right)=0\)
\(\Leftrightarrow2\left(x^2+2x-3\right)+\dfrac{\left(x-2\right)\left(x^2+2x-3\right)}{\sqrt{x^2+2x+6}+3}=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(2+\dfrac{x-2}{\sqrt{x^2+2x+6}+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=0\left(1\right)\\2+\dfrac{x-2}{\sqrt{x^2+2x+6}+3}=0\left(2\right)\end{matrix}\right.\)
- Xét (1): \(x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=2\\x=-3\Rightarrow y=2\end{matrix}\right.\)
- Xét (2): \(\dfrac{x-2}{\sqrt{x^2+2x+6}+3}=-2\)
\(\Leftrightarrow x-2=-2\sqrt{x^2+2x+6}-6\Leftrightarrow2\sqrt{x^2+2x+6}=-x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-4\\4\left(x^2+2x+6\right)=x^2+8x+16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le-4\\3x^2+8=0\end{matrix}\right.\) (vô nghiệm)
Vậy hệ đã cho có 2 cặp nghiệm: \(\left(x;y\right)=\left(1;2\right);\left(-3;2\right)\)
