Question

[Trần An Ly]

giải hpt sau

a) \(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)

b)\(\left\{{}\begin{matrix}x^2+y^2=52\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{12}\end{matrix}\right.\)

Answer 1

[Ánh Lê]

a)

\(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

Đặt x+y = S, xy = P,ta có hệ

\(\left\{{}\begin{matrix}S+P=17\\S^2-P=13\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P=S-17\\S^2-S+4=0\end{matrix}\right.\)

\(S^2-S+4>0\)

=> Hệ phương trình vô nghiệm