Giúp với. Help me

a,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
5x( x-2015) -x+2015=0\\
\Leftrightarrow 5x( x-2015) -( x-2015) =0\\
\Leftrightarrow ( 5x-1)( x-2015) =0\\
\Leftrightarrow \left[ \begin{array}{l l}
5x-1=0 & \\
x-2015=0 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x=\frac{1}{5} & \\
x=2015 & 
\end{array} \right.\\
\end{array}$

b,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{2}{3} x\left( x^{2} -4\right) =0\\
\Leftrightarrow \frac{2}{3} x( x-2)( x+2) =0\\
\Leftrightarrow \left[ \begin{array}{l l}
x=0 & \\
x-2=0 & \\
x+2=0 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x=0 & \\
x=2 & \\
x=-2 & 
\end{array} \right.\\
\end{array}$

c,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
( x+2)^{2} -( x-2)( x+2) =0\\
\Leftrightarrow ( x+2)( x+2-x+2) =0\\
\Leftrightarrow ( x+2) .4=0\\
\Leftrightarrow x+2=0\\
\Leftrightarrow x=-2
\end{array}$

d,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
x+2\sqrt{2} x^{2} +2x^{3} =0\\
\Leftrightarrow x\left( 1+2\sqrt{2} x+2x^{2}\right) =0\\
\Leftrightarrow x\left( 1+\sqrt{2} x\right)^{2} =0\\
\Leftrightarrow \left[ \begin{array}{l l}
x=0 & \\
\sqrt{x} =\frac{-1}{2} & ( vô\ lý)
\end{array} \right.\\
\Leftrightarrow x=0\\
\end{array}$