Question

Giải hệ phương trình :

\(\left\{{}\begin{matrix}x+y=14\\\left(x+y\right)^2-2xy=100\end{matrix}\right.\)

Answer 1

Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} x+y=14\\ 14^2-2xy=100\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+y=14\\ xy=48\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} y=14-x\\ xy=48\end{matrix}\right.\)

\(\Rightarrow x(14-x)=48\)

\(\Leftrightarrow x^2-14x+48=0\)

\(\Leftrightarrow (x-8)(x-6)=0\Rightarrow \left[\begin{matrix} x=8\\ x=6\end{matrix}\right.\)

Khi \(x=8\Rightarrow y=14-x=6\Rightarrow (x,y)=(8,6)\)

Khi \(x=6\Rightarrow y=14-x=8\Rightarrow (x,y)=(6,8)\)

Vậy..........