$Đáp$ $án$ $+$ $Giải$ $thích$ $các$ $bước$ $giải$ $:$

 `1 - ( 2x + 1/2 )^2 = 16/25`

`= ( 2x + 1/2 )^2 = 1 - 16/25`

`= ( 2x + 1/2 )^2 = 25/25 - 16/25`

`= ( 2x + 1/2 )^2 = 9/25`

$\Longrightarrow$ $\left[\begin{matrix} ( 2x + \dfrac{1}{2} )^2 = ( \dfrac{3}{5} )^2 \\ ( 2x + \dfrac{1}{2} )^2 = ( \dfrac{-3}{5} )^2 \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} 2x + \dfrac{1}{2} = \dfrac{3}{5} \\ 2x + \dfrac{1}{2} = \dfrac{-3}{5} \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} 2x = \dfrac{3}{5}  - \dfrac{1}{2} \\ 2x = \dfrac{-3}{5}  - \dfrac{1}{2} \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} 2x = \dfrac{6}{10}  - \dfrac{5}{10} \\ 2x = \dfrac{-6}{10}  - \dfrac{5}{10} \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} 2x = \dfrac{1}{10} \\ 2x = \dfrac{-11}{10} \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} x = \dfrac{1}{10} : 2 \\ x = \dfrac{-11}{10} : 2 \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} x = \dfrac{1}{10} . \dfrac{1}{2} \\ x = \dfrac{-11}{10} . \dfrac{1}{2} \end{matrix}\right.$

$\Longrightarrow$ $\left[\begin{matrix} x = \dfrac{1}{20} \\ x = \dfrac{-11}{20} \end{matrix}\right.$

 $Vậy$ `x in { 1/20 ; {-11}/20 }`