Tìm x, biết:

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

Giải

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

$$\begin{gathered} b. {( x + 2 )}^2-( x - 2 )( x + 2 )=0 \\ (x+2) \left[\right(x+2)-(x-2\left)\right]=0 \\ (x+2)(x+2-x+2)=0 \\ (x+2).4=0 \\ x+2=0 \\ x=-2 \\ Vậy x=-2 \end{gathered}$$

$$\begin{gathered} c. x+2\sqrt{2}{x}^2+2x^3=0 \\ x (1+2\sqrt{2}x+2x^2)=0 \\ x(1^2+2\sqrt{2}x.1+{\left(\sqrt{2}x\right)}^2)=0 \\ x{(1+\sqrt{2}x)}^2=0 \\ Hoặc x=0 \\ Hoặc {(1+\sqrt{2}x)}^2=0\Rightarrow 1+\sqrt{2}x=0\Rightarrow x=-\frac{1}{\sqrt{2}} \\ Vậy x=0, x=-\frac{1}{\sqrt{2}} \end{gathered}$$