Giải phương trình a, 2x(3x-2) - (x+1)(3x-2)=0 b, (x-2)(x^2-3x+5)= x^3-2x^2 c, 1/2(x+1)(3-x)+x=3 d, 3x^2-2x-1=0

a, $\displaystyle 2x( 3x-2) \ -\ ( x+1)( 3x-2) =0$
$\displaystyle \Longrightarrow 6x^{2} -4x-\left( 3x^{2} +x-2\right) =0$
$\displaystyle \Longrightarrow 6x^{2} -4x-3x^{2} -x+2=0$
$\displaystyle \Longrightarrow 3x^{2} -5x+2=0$
$\displaystyle \Longrightarrow ( x-1)( 3x-2) =0$
$\displaystyle \Longrightarrow \left[ \begin{array}{l l}
x-1=0\Longrightarrow x=1 & \\
3x-2=0\Longrightarrow x=\frac{2}{3} & 
\end{array} \right.$
b, $\displaystyle ( x-2)\left( x^{2} -3x+5\right) =\ x^{3} -2x^{2}$
$\displaystyle \Longrightarrow x^{3} -3x^{2} +5x-2x^{2} +6x-10=x^{3} -2x^{2}$
$\displaystyle \Longrightarrow -3x^{2} +11x-10=0$
$\displaystyle \Longrightarrow ( x-2)( 3x-5) =0$
$\displaystyle \Longrightarrow \left[ \begin{array}{l l}
x-2=0\Longrightarrow x=2 & \\
3x-5=0\Longrightarrow x=\frac{5}{3} & 
\end{array} \right.$
d, $\displaystyle 3x^{2} -2x-1=0$
$\displaystyle \Longrightarrow ( x-1)( 3x+1) =0$
$\displaystyle \Longrightarrow \left[ \begin{array}{l l}
x-1=0\Longrightarrow x=1 & \\
3x+1=0\Longrightarrow x=\frac{-1}{3} & 
\end{array} \right.$