\((x-4)(x-5)(x-6)(x-7)=1680 \)
\(\Leftrightarrow\) \((x-4)(x-7)(x-5)(x-6)=1680 \)
\(\Leftrightarrow\) \((x^2-11x+28)(x^2-11x+30)=1680 \)
Đặt \( x^2-11x+28=y\), ta có :
\(y(y+2)=1680 \)
\(\Leftrightarrow\)\(y^2+2y=1680 \)
\(\Leftrightarrow\)\(y^2+2y-1680=0 \)
\(\Leftrightarrow\)\(y^2+42y-40y-1680=0 \)
\(\Leftrightarrow\)\((y+42)(y-40)=0 \)
\(\Leftrightarrow\left[{}\begin{matrix}y=-42\\y=40\end{matrix}\right.\)
Thay vào:
\(\Leftrightarrow\left[{}\begin{matrix}x^2-11x+28=-42\\x^2-11x+28=40\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-11x+70=0\\x^2-11x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-11x=-70\\x^2-11x=12\end{matrix}\right.\)
\(\Leftrightarrow x^2-11x=-70\Rightarrow\)\(x\) vô nghiệm.
\(\Leftrightarrow x^2-11x=12\Rightarrow\)\(\left[{}\begin{matrix}x=-1\\x=12\end{matrix}\right.\)
\(\Rightarrow x=\left\{-1;12\right\}\) .
