Giúp em giải với ạ,em cảm ơn

Bài 1
a,
$\displaystyle x^{3} -2x^{2} +x=x\left( x^{2} -2x+1\right) =x( x-1)^{2}$
b,
$\displaystyle x^{2} -4-4y-y^{2} =x^{2} -\left( 4+4y+y^{2}\right) =x^{2} -( y+2)^{2} =( x-y-2)( x+y+2)$
c,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
3x^{2} -16x+5\\
=3x^{2} -6x-10x+5\\
=3\left( x^{2} -\frac{16}{3} x+\frac{5}{3}\right)\\
=3\left( x^{2} -5x-\frac{1}{3} x+\frac{5}{3}\right)\\
=3\left( x( x-5) -\frac{1}{3}( x-5)\right)\\
=3( x-5)\left( x-\frac{1}{3}\right)
\end{array}$
Bài 2
a,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
3x( x-2) -3x^{2} +12=0\\
\Leftrightarrow 3x^{2} -6x-3x^{2} +12=0\\
\Leftrightarrow -6x+12=0\\
\Leftrightarrow -6x=-12\\
\Leftrightarrow x=2
\end{array}$
b,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
x^{2} -9-( x-3)( 2x-5) =0\\
\Leftrightarrow x^{2} -9-\left( 2x^{2} -6x-5x+15\right) =0\\
\Leftrightarrow x^{2} -9-2x^{2} +6x+5x-15=0\\
\Leftrightarrow -x^{2} +11x-24=0\\
\Leftrightarrow x^{2} -11x+24=0\\
\Leftrightarrow ( x-8)( x-3) =0\\
\Leftrightarrow \left[ \begin{array}{l l}
x-8=0 & \\
x-3=0 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x=8 & \\
x=3 & 
\end{array} \right.
\end{array}$
c,
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
x( 3x-16) =-5\\
\Leftrightarrow 3x^{2} -16x+5=0\\
\Leftrightarrow 3( x-5)\left( x-\frac{1}{3}\right) =0\\
\Leftrightarrow \left[ \begin{array}{l l}
x-5=0 & \\
x-\frac{1}{3} =0 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x=5 & \\
x=\frac{1}{3} & 
\end{array} \right.
\end{array}$