(2x-3)^2-16=0 ; (2x-3+4)*(2x-3-4)=0,所以满足其中之一为0即可,x=-1/2或者7/2
(2x-3)^2-16
=(2x-3+4)(2x-3-4)
=(2x+1)(2x-7)=0
2x+1=0或2x-7=0
x1=-1/2
x2=7/2
平方差
(2x-3)^2-4^2=0
(2x-3+4)(2x-3-4)=0
(2x+1)(2x-7)=0
x=-1/2,x=7/2
由题得4x^2-12x+9-16=0
4x^2-12x-7=0
(2x-7)(2x+1)=0
x=7/2 或 x=-1/2
(2X-3)²-16=0
(2X-3-4)*(2X-3+4)=0
(2X-7)*(2X+1)=0
2X-7=0 2X+1=0
X1=7/2 X2=-1/2