a/ \(\left\{{}\begin{matrix}xy+1+x+y=10\\\left(x+y\right)\left(xy+1\right)=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=10\\ab=1\end{matrix}\right.\)
Theo Viet đảo, a và b là nghiệm:
\(t^2-10t+1=0\) \(\Rightarrow\left[{}\begin{matrix}t=5+2\sqrt{6}\\t=5-2\sqrt{6}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=5+2\sqrt{6}\\xy=4-2\sqrt{6}\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm:
\(t^2-\left(5+2\sqrt{6}\right)t+4-2\sqrt{6}=0\) (bấm máy, số xấu quá)
TH2: \(\left\{{}\begin{matrix}x+y=5-2\sqrt{6}\\xy=4+2\sqrt{6}\end{matrix}\right.\)
Ta có \(\left(5-2\sqrt{6}\right)^2-4\left(4+2\sqrt{6}\right)=33-28\sqrt{6}< 0\) nên vô nghiệm
b/ \(\left\{{}\begin{matrix}x^4+y^4=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)^2-2x^2y^2=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+y^2=a>0\\xy=b\end{matrix}\right.\) với \(a\ge2b\) hệ trở thành:
\(\left\{{}\begin{matrix}a^2-2b^2=97\\ab=78\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=97\\b=\frac{78}{a}\end{matrix}\right.\)
\(\Rightarrow a^2-2\left(\frac{78}{a}\right)^2=97\)
\(\Leftrightarrow a^4-97a^2-12168=0\Rightarrow\left[{}\begin{matrix}a^2=169\\a^2=-72\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=13\Rightarrow b=6\\a=-13< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\xy=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\y=\frac{6}{x}\end{matrix}\right.\)
\(\Rightarrow x^2+\frac{36}{x^2}=13\Leftrightarrow x^4-13x^2+36=0\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\x^2=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\Rightarrow y=2\\x=-3\Rightarrow y=-2\\x=2\Rightarrow y=3\\x=-2\Rightarrow y=-3\end{matrix}\right.\)
