Tìm x,biết 

$$\begin{gathered} a) x^3-\frac{1}{4}=0; \\ b){(2x-1)}^2-{(x+3)}^2=0; \\ c\left)x^2\right(x-3)+12-4x=0; \end{gathered}$$

Giải:

$$\begin{gathered} a) x^3-\frac{1}{4}=0 \\ \Rightarrow x\left(x^2-{\left(\frac{1}{2}\right)}^2\right)=0 \\ \Rightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0 \\ Hoặc x=0 \\ Hoặc x-\frac{1}{2}=0 \Rightarrow x=\frac{1}{2} \\ Hoặc x+\frac{1}{2}=0 \Rightarrow x=-\frac{1}{2} \\ Vậy x=0; x=-\frac{1}{2}; x=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} b){(2x-1)}^2-{(x+3)}^2=0 \\ [(2x - 1) - (x + 3)\left]\right[(2x - 1) + (x + 3)] = 0 \\ (2x - 1 - x - 3\left)\right(2x - 1 + x + 3) = 0 \\ (x - 4\left)\right(3x + 2) = 0 \\ Hoặc x-4=0\Rightarrow x=4 \\ Hoặc 3x+2=0\Rightarrow 3x=2\Rightarrow x=-\frac{2}{3} \\ Vậy x=4; x=-\frac{2}{3} \end{gathered}$$

$$\begin{gathered} c\left)x^2\right(x-3)+12-4x=0 \\ {x}^2(x – 3) - 4(x -3)= 0 \\ (x - 3\left)\right(x^2- 2^2) = 0 \\ (x - 3\left)\right(x - 2\left)\right(x + 2) = 0 \\ Hoặc x-3 \Rightarrow x=3 \\ Hoặc x-2 \Rightarrow x=2 \\ Hoặc x+2 \Rightarrow x=-2 \\ Vậy x=3; x=2; x=-2. \end{gathered}$$