Giai phuong trinh a,x^2-x-(3x-3)=0 b,3x-14=2x(x-5) c,(x-1)(5x+3)=(3x-8)(x-1) d,(x-2)(2x+3)-(2-x)(4-3x)=0

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

   b. $3x – 14 = 2(x-5)$
   ⇔ $3x – 14 -2(x-5)=0$
   ⇔ $3x – 14 -2x + 10 = 0$ 
   ⇔ $x = 4$
   $S =$ { $4$ }

   c. $( x – 1 )( 5x+3)=(3x-8)(x-1)$
   ⇔$(x-1)(5x+3)-(3x-8)(x-1)=0$
   ⇔$ ( x-1)(5x+3-3x+8)=0$
   ⇔$(x-1)(2x+11)=0$
   ⇔\(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
   ⇔\(\left[ \begin{array}{l}x=1\\x=\dfrac{-11}{2}\end{array} \right.\) 

   $S =$ { $1; \dfrac{-11}{2}$ }

   d. $(x-2)(2x+3)-(2-x)(4-3x)=0$
   ⇔$(x-2)(2x+3)+(x-2)(4-3x)=0$
   ⇔$(x-2)(2x+3+4-3x)=0$
   ⇔$(x-2)(7-x)=0$
   ⇔\(\left[ \begin{array}{l}x-2=0\\7-x=0\end{array} \right.\)
   ⇔\(\left[ \begin{array}{l}x=2\\x=7\end{array} \right.\)

   $S =$ { $2; 7$ }

    

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