\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Rightarrow\left(x^2+4x-21\right)\left(x^2+4x-5\right)-297=0\)
Đặt \(t=x^2+4x-5\) ta có:
\(\left(t-16\right)t-297=0\)\(\Rightarrow t^2-16t-297=0\)
\(\Rightarrow t^2-27t+11t-297=0\)
\(\Rightarrow t\left(t-27\right)+11\left(t-27\right)=0\)
\(\Rightarrow\left(t+11\right)\left(t-27\right)=0\)\(\Rightarrow\left[\begin{matrix}t=-11\\t=27\end{matrix}\right.\)
*)Xét \(t=-11\Rightarrow x^2+4x-5=-11\)
\(\Rightarrow x^2+4x+6=0\Rightarrow\left(x+2\right)^2+2>0\left(loai\right)\)
*)Xét \(t=27\Rightarrow x^2+4x-5=27\)
\(\Rightarrow x^2+4x-32=0\Rightarrow\left(x+8\right)\left(x-4\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-8\\x=4\end{matrix}\right.\)
