\(\Leftrightarrow2\left(x^2+2x+1\right)^2-\left(2x-1\right)\left(3x^2+10x+1\right)=0\)
Đặt \(\left\{{}\begin{matrix}2x^2+6x+1=a\\x^2+4x=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=x^2+2x+1\\2b-a=2x-1\\a+b=3x^2+10x+1\end{matrix}\right.\)
Phương trình trở thành:
\(2\left(a-b\right)^2-\left(2b-a\right)\left(a+b\right)=0\)
\(\Leftrightarrow2a^2-4ab+2b^2-\left(2b^2+ab-a^2\right)=0\)
\(\Leftrightarrow3a^2-5ab=0\)
\(\Leftrightarrow a\left(3a-5b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\\3a-5b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+6x+1=0\\6x^2+18x+3-5x^2-20x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+6x+1=0\left(casio\right)\\x^2-2x+3=0\left(vn\right)\end{matrix}\right.\)
