Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}x+y=1\\x^2-y^2=11\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}x+y=1\\x^2-y^2=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\\left(x-y\right)\left(x+y\right)=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\\left(1-y-y\right)\left(1-y+y\right)=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\1-2y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-5\right)\\y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=-5\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm là \(\left(6;-5\right)\)