Trừ vế cho vế:
\(x^2-y^2+y-x=0\)
\(\Rightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=1-x\end{matrix}\right.\)
Thế vào pt đầu:
\(\Rightarrow\left[{}\begin{matrix}x^2+x=1\\x^2+1-x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=\dfrac{-1\pm\sqrt{5}}{2}\\x=0;y=1\\x=1;y=0\\\end{matrix}\right.\)
Đúng 0
Bình luận (0)
