Tìm x: a)3x(x-2)-x+2=0 b)4x(x-3)-2x+6=0 c)2x^3+4x=0 d)(x-2)^2-(x-3)(x+3)=6 e)4(x-3)^6-(2x-1)(2x+1)=10 f)x^2+5x+6=0 g)x^3-9x^2+20x=0

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1. $a) 3x(x-2)-x+2=0$

   ⇔$3x(x-2)-(x-2)=0$

   ⇔$(x-2)(3x-1)=0$

   ⇔\(\left[ \begin{array}{l}x-2=0\\3x-1=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=2\\x=\frac{1}{3}\end{array} \right.\)

   $b)4x(x-3)-2x+6=0$

   ⇔$4x(x-3)-2(x-3)=0$

   ⇔$(x-3)(4x-2)=0$

   ⇔\(\left[ \begin{array}{l}x-3=0\\4x-2=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=3\\x=\frac{1}{2}\end{array} \right.\) 

   $c) 2x^{3}-4x=0$ 

   ⇔$2x(x^{2}-4)=0$ 

   ⇔$2x(x-2)(x+2)=0$ 

   ⇔$x=0$ $hoặc$  \(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\) 

   ⇔$x=0$ $hoặc$ \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) 

   $d)(x-2)^2-(x-3)(x+3)=9$

   ⇔$(x-2)^2-x^2+9=9$

   ⇔$(x-2)^2-x^2=0$

   ⇔\frac{

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