a/ `( x - 4)( x- 5 )( x - 6 )( x - 7 ) = 1680`
`\Leftrightarrow ( x^2 - 11x + 28 )( x^2 -11x + 30 ) = 1680`
Đặt `x^2 -11x +29 = t`, khi đó:
`( t - 1 )( t + 1 ) = 1680`
`\Leftrightarrow t^2 - 1681=0`
`\Leftrightarrow ( t + 41 )( t - 41 ) = 0`
`\Leftrightarrow` \(\left[ \begin{array}{l}t=-41\\t=41\end{array} \right.\)
`\Leftrightarrow` \(\left[ \begin{array}{l}x^2 -11x + 29 = -41 ( vn )\\x^2 - 11x + 29 = 41 \end{array} \right.\)
`\Leftrightarrow` \(\left[ \begin{array}{l}x+1=0\\x-12=0\end{array} \right.\)
`\Leftrightarrow` \(\left[ \begin{array}{l}x=-1\\x=12\end{array} \right.\)
`S =` { `-1;12` }
b/ `2x^3 + 5x^2 = 7x`
`\Leftrightarrow x( 2x^2 - 2x + 7x - 7 ) = 0`
`\Leftrightarrow x[ 2x( x - 1 ) + 7( x-1) ] = 0`
`\Leftrightarrow x( x - 1 )( 2x + 7 ) = 0`
`\Leftrightarrow` \(\left[ \begin{array}{l}x=0\\x=1\\x = \dfrac{-7}{2}\end{array} \right.\)
`S =` { `\frac{-7}{2}; 0 ; 1` }
