Bài :tìm x a,3x(4x mũ 2-1)=0 b,(x+5)mũ 2-(x+5)(x-2)=0 c,2x mũ 2-4x mũ 2+2x=0 d,(x+1) mũ 2-(2x+3) mũ 2=0

a, $3x(4x^2-1)=0$

    $⇔3x(2x-1)(2x+1)=0$

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

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

Vậy $S=\bigg\{0;\pm\dfrac{1}{2}\bigg\}$

b, $(x+5)^2-(x+5)(x-2)=0$

    $⇔(x+5)(x+5-x+2)=0$

    $⇔(x+5).7=0$

    $⇔x=-5$

Vậy $S=\{-5\}$

c, $2x^2-4x^2+2x=0$

   $⇔-2x^2+2x=0$

   $⇔-2x(x-1)=0$

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

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

Vậy $S=\{0;1\}$

d, $(x+1)^2-(2x+3)^2=0$

    $⇔(x+1-2x-3)(x+1+2x+3)=0$

    $⇔(-x-2)(3x+4)=0$

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

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

Vậy $S=\bigg\{-2;\dfrac{-4}{3}\bigg\}$