Question

giải hệ phương trình 

\(\left\{{}\begin{matrix}x^{2^{ }}-5y+4=0\\y^2-5x+4=0\end{matrix}\right.\)

Answer 1

Trừ vế cho vế:

\(x^2-y^2+5\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+5\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+5\right)=0\)

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

Thế vào pt đầu:

\(\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5\left(-x-5\right)+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2+5x+29=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=4\Rightarrow y=4\end{matrix}\right.\)