$\text {1,}$
`(x+3)^2.(x-4) =0`
`⇔`\(\left[ \begin{array}{l}(x+3)^2=0\\x-4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x+3=0\\x=4\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-3\\x=4\end{array} \right.\)
$\text {Vậy S∈}$ `{-3;4}`
$\text {2,}$
`(2x-4)^2.(2x-6) =0`
`⇔`\(\left[ \begin{array}{l}(2x-4)^2=0\\2x-6 =0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x-4 =0\\2x=6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
$\text {Vậy S∈}$ `{2;3}`
$\text {3,}$
`(4x+8)^2.(2x-6) =0`
`⇔`\(\left[ \begin{array}{l}(4x+8)^2 =0\\2x-6 =0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}4x+8=0\\2x=6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
$\text {Vậy S∈}$ `{-2;3}`
$\text {4,}$
`(x^2+1)(x-1)^2(3x-1) =0`
`⇔`\(\left[ \begin{array}{l}x^2+1=0\\(x-1)^2=0\\3x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x^2=-1 (Không TM)\\x-1=0\\3x=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x^2=-1 (Không TM)\\x=1\\x=\frac{1}{3} \end{array} \right.\)
$\text {Vậy S∈}$ `{1;\frac{1}{3}}`
$\text {@lamtung2007}$
