\(\left(x^2+2x+8\right)=y^2.\)
\(\left(x+1\right)^2-y^2=-7\)
\(\left(x-y+1\right)\left(x+y+1\right)=-7\)
+\(\hept{\begin{cases}x+y+1=7\\x-y+1=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=6\\x-y=-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}}\)
+\(\hept{\begin{cases}x+y+1=1\\x-y+1=-7\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\x-y=-8\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-4\left(L\right)\\\end{cases}}}\)
Vậy x =2
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x=2 đúng đấy bạn
\(x^2+2x+8=2^2+2.2+8=16=4^2\)(đúng)