(2x-1)^+(1-2x)-6=0
即:(1-2x)^+(1-2x)-6=0
由十字相乘法: (1-2x+3)(1-2x-2)=0
所以,(4-2x)(-2x-1)=0,解得:x1=2,x2=-1/2
(2x-1)²-(2x-1)-6=0
[(2x-1)+2][(2x-1)-3]=0即(2x+1)(2x-4)=0
2x+1=0或2x-4=0
∴x= -1/2或x=2
:(2x-1)平方+(1-2x)-6=0
因式分解得
(2x-1-3)(2x-1+2)=0
就是2x-1-3=0或2x-1+2=0
解得x=2或x=-1/2
4x²-4x+1+1-2x-6=0
4x²-6x-4=0
都除2
2x²-3x-2=0
(2x+1)(x-2)=0
x1=-1/2
x2=2
不懂可追问,有帮助请采纳,谢谢!
(2x-1)²-(2x-1)-6=0
(2x-1-3)(2x-1+2)=0
(2x-4)(2x+1)=0
x=2,x=-1/2