Giải các phương trình sau: a) x² – 5x + 6 = 0 b) 2x³ + 6x² = x² + 3x c) (3x – 1) (x² + 2) = (3x – 1) (7x – 10)

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1. a, $x^2-5x+6=0$

   $⇔x^2-2x-3x+6=0$

   $⇔x(x-2)-3(x-2)=0$

   $⇔(x-2)(x-3)=0$

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

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

   Vậy $S=\{2;3\}$

   b, $2x^3+6x^2=x^2+3x$

   $⇔2x^3+5x^2-3x=0$

   $⇔x(2x^2+5x-3)=0$

   $⇔x(2x^2+6x-x-3)=0$

   $⇔x(x+3)(2x-1)=0$

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

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

   Vậy $S=\bigg\{-3;0;\dfrac{1}{2}\bigg\}$

   c, $(3x-1)(x^2+2)=(3x-1)(7x-10)$

   $⇔(3x-1)(x^2+2)-(3x-1)(7x-10)=0$

   $⇔(3x-1)(x^2+2-7x+10)=0$

   $⇔(3x-1)(x^2-7x+12)=0$

   $⇔(3x-1)(x^2-3x-4x+12=0)=0$

   $⇔(3x-1)(x-3)(x-4)=0$

   \(⇔\left[ \begin{array}{l}3x-1=0\\x-3=0\\x-4=0\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=3\\x=4\end{array} \right.\)

   Vậy $S=\bigg\{\dfrac{1}{3};3;4\bigg\}$

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2. $a) x^2 -5x +6=0$

   $\to x^2 -2x -3x+6=0$

   $\to x(x-2)-3(x-2)=0$

   $\to (x-2)(x-3)=0$

   Nếu:

   $x-2=0 \to x=2$

   $x-3=0 \to x=3$

   Vậy $S={2;3}$

   $b)2x^3+6x^2=x^2+3x$

   $\to 2x^2(x+3)=x(x+3)$

   $\to 2x^2(x+3)-x(x+3)=0$

   $\to (x+3)(2x^2-x)=0$

   $\to x(2x-1)(x+3)=0$

   Nếu: 

   $x=0 \to x=0$

   $2x-1=0 \to x=\dfrac{1}{2}$

   $x+3=0 \to x=-3$

   Vậy $S={-3;0;\dfrac{1}{2}}$

   $c)(3x-1)(x^2+2)=(3x-1)(7x-10)$

   $\to (3x-1)(x^2+2)-(3x-1)(7x-10)=0$

   $\to (3x-1)(x^2+2-7x+10)=0$

   $\to (3x-1)(x^2-7x+12)=0$

   Nếu :

   $x^2-7x+12=0$

   $\to x^2 -3x-4x+12=0$

   $\to x(x-3)-4(x-3)=0$

   $\to (x-3)(x-4)=0$

   $\to x=3;x=4$

   $3x-1=0$

   $\to x=\dfrac{1}{3}$

   Vậy $S={\dfrac{1}{3};3;4}$

    

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