a, $(x-1)(3x-6)=0$
$⇔3(x-1)(x-2)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
Vậy S={1;2}
b, $(2x+5)(1-3x)=0$
$⇔$\(\left[ \begin{array}{l}2x+5=0\\1-3x=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=-5/2\\x=1/3\end{array} \right.\)
Vậy S={-5/2;1/3}
c, $(x+1)(2x-3)(3x-5)=0$
⇔\(\left[ \begin{array}{l}x+1=0\\2x-3=0\\3x-5=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=-1\\x=3/2\\x=5/3\end{array} \right.\)
Vậy S={-1;3/2;5/3}
d, *Câu này mình không rõ đề, nên không làm được, mờ quá
e, $(3z-2)^2(z+1)(z-2)=0$
$⇔$\(\left[ \begin{array}{l}3z-2=0\\z+1=0\\z-2=0\end{array} \right.\)=>\(\left[ \begin{array}{l}z=2/3\\z=-1\\z=2\end{array} \right.\)
Vậy S={2/3;-1;2}
f, $(8-x)^2(3x+6)=0$
$⇔3(8-x)^2(x+2)=0$
$⇔$\(\left[ \begin{array}{l}8-x=0\\x+2=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\)
Vậy S={8;2}
g, $(1/2x+1)^3(3x+1)^2=0$
$⇔$\(\left[ \begin{array}{l}1/2x+1=0\\3x+1=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=-2\\x=-1/3\end{array} \right.\)
Vậy S={-2;-1/3}
