Đặt \(x^2+2x=a\), pt trở thành:
\(a^2+2a=15\Leftrightarrow a^2+2a-15=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\\a=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\\x^2-2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\x\in\varnothing\left[x^2-2x+5=\left(x-1\right)^2+4>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Ta có: \(\left(x^2-2x\right)^2+2\left(x^2-2x\right)=15\)
\(\Leftrightarrow\left(x^2-2x\right)^2+2\left(x^2-2x\right)-15=0\)
\(\Leftrightarrow\left(x^2-2x+5\right)\left(x^2-2x-3\right)=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)