a) Ta có:
\(\eqalign{
& {x^3} - {1 \over 4}x = 0 \cr& x\left( {{x^2} - {1 \over 4}} \right) = 0 \cr
& x\left( {{x^2} - {{\left( {{1 \over 2}} \right)}^2}} \right) = 0 \cr
& x\left( {x - {1 \over 2}} \right)\left( {x + {1 \over 2}} \right) = 0 \cr
& \Rightarrow \left[ \matrix{
x = 0 \hfill \cr
{x - \dfrac{1}{2}} = 0 \Rightarrow x = \dfrac{1}{2} \hfill \cr
{x + \dfrac{1}{2}} = 0 \Rightarrow x = - \dfrac{1}{2} \hfill \cr} \right. \cr} \)
Vậy \(x=0,x = \dfrac{1}{2},x = - \dfrac{1}{2}\)
c) Ta có:
\(\eqalign{
& {x^2}(x - 3) + 12 - 4x = 0 \cr
& {x^2}(x - 3) + \left( { - 4x + 12} \right) = 0\cr&{x^2}(x - 3) - 4(x - 3) = 0 \cr
& (x - 3)({x^2} - 4) = 0 \cr
& (x - 3)(x - 2)(x + 2) = 0 \cr
& \Rightarrow \left[ \begin{gathered}x - 3 = 0 \hfill \\x - 2 = 0 \hfill \\x + 2 = 0 \hfill \\ \end{gathered} \right.\Rightarrow \left[ \matrix{x = 3 \hfill \cr x = 2 \hfill \cr x = - 2 \hfill \cr} \right. \cr} \)
Vậy \( x=3,x=2,x=-2\)
