Question

(x-2)的平方-9(x+1)的平方=0，这个方程怎么用开平方法解?

Answer 1

(x-2)^2 - 9(x+1)^2 = 0
(x-2)^2 - [3(x+1)]^2 = 0
由平方差公式得：[(x-2) + 3(x+1)][(x-2) - 3(x+1)] = 0
整理得：(4x + 1)(-2x - 3) = 0
所以x = -1/4 或 -3/2

Answer 2

解：(x-2)²=9(x+1)²
(x-2)²=[3(x+1)]²
x-2=±[3(x+1)]
x-2=3(x+1),或x-2=-3(x+1)
x-2=3x+3,或x-2=-3x-3
3x-x=-2-3,或x+3x=-3+2
2x=-5或4x=-1
x=-5/2或x=-1/4

Answer 3

(x-2)^2 - 9(x+1)^2 = 0
(x-2)^2 =9(x+1)^2
|x-2 |=3|(x+1) |
x-2 =3(x+1) 或 x-2 =-3(x+1)
解得 x= -5/2 或x=-1/4