Question

(x-2)二次方=(2x+3)二次方 用分解因式法怎么解

Answer 1

(2x+3)²-(x-2)²=0
(2x+3+x-2)(2x+3-x+2)=0
(3x+1)(x+5)=0
x=-1/3,x=-5

Answer 2

A-B
=a²-a+17/4-2a-1
=a²-3a+13/4
=a²-3a+9/4+1
=(a-3/2)²+1≥1>0
所以 A>B
f(x)定义域是[0,1]
则f(x²+1)有0<=x²+1<=1
-1<=x²<=0
只有x²=0成立
x=0
所以定义域是{0}