Giải phương trình: a,(2x-1) cộng(x-3).(2x-1) bằng 0 b,2(x-5).(x cộng 2) bằng x^2 – 5x c,(2x cộng 1).(1-x) cộng 2x bằng 2 d,x^2 – 5x cộng 6 bằng 0 Giúp

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

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

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