\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\)
\(\Leftrightarrow\)\(\left(x^2+4x-5\right)\left(x^2+4x-21\right)-297=0\)
Đặt \(x^2+4x-5=t\) ta có:
\(t\left(t-16\right)-297=0\)
\(\Leftrightarrow\)\(t^2-16t+64-361=0\)
\(\Leftrightarrow\) \(\left(t-8\right)^2-361=0\)
\(\Leftrightarrow\)\(\left(t-8-19\right)\left(t-8+19\right)=0\)
\(\Leftrightarrow\)\(\left(t-27\right)\left(t+11\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}t-27=0\\t+11=0\end{cases}}\)
Thay trở lại ta được: \(\orbr{\begin{cases}x^2+4x-32=0\\x^2+4x+6=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-4\right)\left(x+8\right)=0\\\left(x+2\right)^2+2=0\left(L\right)\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)
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