a, ( 2x-1) + ( x-3).( 2x-1)= 0
⇔ ( 2x-1).( x-3+1)= 0
⇔ ( 2x-1).( x-2)= 0
⇔ \(\left[ \begin{array}{l}2x-1=0\\x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0,5\\x=2\end{array} \right.\)
b, 2( x-5).( x + 2)= x² - 5x
⇔ 2( x-5).( x + 2)= x.( x-5)
⇔ ( x-5).( 2x+4-x) = 0
⇔ ( x-5).( x+4)=0
⇔ \(\left[ \begin{array}{l}x-5=0\\x+4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5\\x=-4\end{array} \right.\)
c, ( 2x+1).( 1-x)+2x=2
⇔ ( 2x+1).( 1-x)+2x-2=0
⇔ ( 2x+1).( 1-x)+2.( x-1)=0
⇔ ( x-1).( -2x-1+2)=0
⇔ ( x-1).( -2x+1)
⇔ \(\left[ \begin{array}{l}x-1=0\\-2x+1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=0,5\end{array} \right.\)
d, x²-5x+6=0
⇔ x²-3x-2x+6=0
⇔ x.( x-3)-2.( x-3)=0
⇔ ( x-3).( x-2)= 0
⇔ \(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
