Question

Giải hệ phương trình \(\left\{{}\begin{matrix}a^2+b^2=1\\a+b+ab=1\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}a^2+b^2=1\\a+b+ab=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=1\\a+b+ab=1\end{matrix}\right.\) Đặt \(\left\{{}\begin{matrix}a+b=S\\a.b=P\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}S^2-2P=1\\S+P=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}S^2-2\left(S-1\right)=1\\P=1-S\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}S=1\\P=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\a.b=0\end{matrix}\right.\) => a,b là NPT:a²-a=0 \(\Rightarrow a_1=0;a_2=1\)

=>N (a;b)của hệ là :{(0;1);(1;0)}