因式分解法解方程(x+1)平方=2(x+1)

分类: 作业答案 • 2021-12-20 09:03:14

问题描述：

答

(x + 1)² - 2(x + 1) = 0
(x + 1)(x + 1 - 2) = 0
(x + 1)(x - 1) = 0
x1 = 1，x2 = -1

答

(x+1)^2=2(x+1)
(x+1)^2-2(x+1)=0
(x+1)(x+1-2)=0
(x+1)(x-1)=0
x+1=0
x=-1
x-1=0
x=1

答

﹙x＋1﹚²－2﹙x＋1﹚＝0
﹙x＋1﹚﹙x－1﹚＝0
x1＝﹣1
x2＝1.

答

(x + 1)² = 2(x + 1)
(x + 1)² - 2(x + 1) = 0
(x + 1)(x + 1 - 2) = 0
(x + 1)(x - 1) = 0
x1 = 1，x2 = -1