`a)` `x^3 + x = 0`

`=> x.(x^2+1) =0`

`=>` $\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.$ 

`=>` $\left[ \begin{array}{l}x=0\\x^2=-1\end{array} \right.$ 

`=> x =0`

`b)` `(x-2024) - (x-2024) = 0`

`=> x - 2024 - x + 2024 =0`

`=> 0 = 0`

Vậy với `(x-2024) - (x-2024) = 0 ∀x`

`c) x^3 + 2x^2 + x = 0`

`=> x.(x^2+2x+1) = 0`

`=> x.(x-1)^2=0`

`=>` $\left[ \begin{array}{l}x=0\\(x-1)^2=0\end{array} \right.$ 

`=>` $\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.$ 

`=>` $\left[ \begin{array}{l}x=0\\x=1\end{array} \right.$ 

`d)` `15x^2 + 3x= 0`

`=> 3x.(5x+1)=0`

`=>` $\left[ \begin{array}{l}3x=0\\5x+1=0\end{array} \right.$ 

`=>` $\left[ \begin{array}{l}x=0\\5x=-1\end{array} \right.$ 

`=>` $\left[ \begin{array}{l}x=0\\x=-1/5\end{array} \right.$ 

`e)` `x^3 - 25x =0`

`=> x.(x^2-25)=0`

`=> x.(x-5)^2=0`

`=>` $\left[ \begin{array}{l}x=0\\(x-5)^2=0\end{array} \right.$ 

`=>` $\left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.$ 

`=> $\left[ \begin{array}{l}x=0\\x=5\end{array} \right.$