Ta có:\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)\left(x-3\right)\left(x+7\right)=297\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\)
Đặt \(x^2+4x-5=t\) thì \(t\left(t-16\right)=297\)
\(\Leftrightarrow t^2-16t-297=0\Leftrightarrow t^2-27t+11t-297=0\)
\(\Leftrightarrow t\left(t-27\right)+11\left(t-27\right)=0\Leftrightarrow\left(t+11\right)\left(t-27\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=-11\\t=27\end{cases}}\)
Với \(t=-11\) thì \(x^2+4x-5=-11\Leftrightarrow x^2+4x+6=0\Leftrightarrow x^2+4x+4+2=0\)
\(\Leftrightarrow\left(x+2\right)^2+2=0\)(vô lí)
Với \(t=27\) thì \(x^2+4x-5=27\Leftrightarrow x^2+4x-32=0\Leftrightarrow x^2-4x+8x-32=0\)
\(\Leftrightarrow x\left(x-4\right)+8\left(x-4\right)=0\Leftrightarrow\left(x+8\right)\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-8\\x=4\end{cases}}\)
Tập nghiệm của pt \(S=\left\{-8,4\right\}\)
\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)
\(\Leftrightarrow\left(x^2+5x-x-5\right)\left(x^2+7x-3x-21\right)=297\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\)
Đặt \(x^2+4x-13=m\)
Ta có : \(\left(m+8\right)\left(m-8\right)=297\)
\(\Leftrightarrow m^2-8^2=297\)
\(\Leftrightarrow m^2=361\)
\(\Leftrightarrow m=\pm19\)
+) Với m = 19 ta có : \(x^2+4x-13=19\)
\(\Leftrightarrow x^2+4x-32=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(8x-32\right)=0\)
\(\Leftrightarrow x\left(x-4\right)+8\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)
+) Với m = -19 ta có : \(x^2+4x-13=-19\)
\(\Leftrightarrow x^2+4x+6=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)+2=0\)
\(\Leftrightarrow\left(x+2\right)^2+2=0\)
\(\Leftrightarrow\left(x+2\right)^2=-2\) ( vô lí )
Vậy phương trình có tập nghiệm \(S=\left\{4;-8\right\}\)
