Lời giải
Khử trị tuyệt đối
\(\left|\left(y-x-1\right)^2+x-2\right|+4=2x-\left|\left(x-1\right)\left(x-2\right)\right|\)
VT >= 4 =>để có nghiệm VP >=4
=> x>=2
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-2\right)\ge0\\\left(y-x-1\right)^2+\left(x-2\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\left(y-x-1\right)^2+x\right|=\left(y-x-1\right)^2+\left(x-2\right)\\\left|\left(x-1\right)\left(x-2\right)\right|=\left(x-1\right)\left(x-2\right)\end{matrix}\right.\)
Phương trình tương đương hệ
\(\left\{{}\begin{matrix}x\ge2\left(1\right)\\\left(x-y+1\right)^2+\left(x-2\right)+4=2x-\left(x-1\right)\left(x-2\right)\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(x-y+1\right)^2=\left(x-2\right)-\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(x-y+1\right)^2=\left(x-2\right)\left[1-\left(x-1\right)\right]=-\left(x-2\right)^2\)
\(\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)=0\\x-y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Kết luận
(x,y) =(2,3) là nghiệm duy nhất