Question

giải hệ sau \(\left\{{}\begin{matrix}3x^2-2x-5+2x\sqrt{x^2+1}=2\left(y+1\right)\sqrt{y^2+2y+2}\\x^2+2y^2-2x+4y-3=0\end{matrix}\right.\)

Answer 1

Trừ vế cho vế:

\(\Rightarrow2x^2-2-2y^2-4y+2x\sqrt{x^2+1}=2\left(y+1\right)\sqrt{y^2+2y+2}\)

\(\Leftrightarrow x^2+x\sqrt{x^2+1}=\left(y+1\right)^2+\left(y+1\right)\sqrt{\left(y+1\right)^2+1}\)

Xét hàm \(f\left(t\right)=t^2+t\sqrt{t^2+1}\)

\(f'\left(t\right)=\frac{\left(\sqrt{t^2+1}+t\right)^2}{\sqrt{t^2+1}}>0\) ;\(\forall t\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow x=y+1\)

Thay xuống pt dưới:

\(\left(y+1\right)^2+2y^2-2\left(y+1\right)+4y-3=0\Leftrightarrow...\)