Question

[Đinh Đức Châu]

tìm x,y thoả mãn x ²+x-y²  =0

^{cíu mình với}

Answer 1

[Nguyễn Ngọc Anh Minh]

\(\Leftrightarrow x\left(x+1\right)=y^2\)

+ Nếu \(y=0\)

\(\Rightarrow x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

+ Nếu \(y\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}x=y^m\\x+1=y^n\end{matrix}\right.\)

\(\Rightarrow x\left(x+1\right)=y^m.y^n=y^{m+n}=y^2\Rightarrow m+n=2\) (1)

Ta có

\(y^n-y^m=\left(x+1\right)-x=1\)

\(\Leftrightarrow y^n\left(1-y^{m-n}\right)=1.1\)

\(\Rightarrow\left\{{}\begin{matrix}y^n=1\\y^{m-n}=0\end{matrix}\right.\) (2)

Kết hợp (1) và (2)

\(\Rightarrow\left\{{}\begin{matrix}y^n=1\\y^{m-n}=0\\m+n=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=0\\y=0\\m=2\end{matrix}\right.\) mâu thuẫn với đk \(y\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\y=0\end{matrix}\right.\)