Giải:
pt\(\Leftrightarrow\) y4+4y3+6y2+4y+1+y4 = x2+2x+1+x2
\(\Leftrightarrow\) 2y4+4y3+6y2+4y = 2x2+2x
\(\Leftrightarrow\) y4+2y3+3y2+2y = x2+x
\(\Leftrightarrow\) y4+2y3+3y2+2y +1 = x2+x+1
\(\Leftrightarrow\) (y2+y+1)2 = x2+x+1
Cách 1:Đưa về pt tích số
Đặt k = y2+y+1 ( k\(\in\) N*)
\(\Rightarrow\) k2 = x2+x+1
\(\Leftrightarrow\) 4k2 = 4x2+4x+4
\(\Leftrightarrow\) 4k2 = (2x+1)2 +3
\(\Leftrightarrow\) (2x+1)2 - 4k2 = -3
\(\Leftrightarrow\) (2x+1-2k)(2x+1+2k) = -3=-1.3=-3.1
( do 2x+1-2k < 2x+1+2k
Xét từng trường hợp
TH1: \(\left\{{}\begin{matrix}2x+1-2k=-1\\2x+1+2k=3\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0\\k=1\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}2x+1-2k=-3\\2x+1+2k=1\end{matrix}\right.\)
\(\Rightarrow\)\(\left\{{}\begin{matrix}x=-1\\k=1\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)
Cách 2: Sử dụng nguyên lí kẹp
Do x\(\ge\) 0 \(\Rightarrow\) x2 \(\le\) x2+x+1 \(\le\) (x+1)2
\(\Leftrightarrow\) x2 \(\le\) x2+x+1 \(\le\) x2+2x+1
Xét x2 \(\le\) x2+x+1 \(\Rightarrow\) x = -1 \(\Rightarrow\) y=0
Xét x2+x+1 \(\le\) x2+2x+1 \(\Rightarrow\) x = 0 \(\Rightarrow\) y = 0
