已知不等式组X+3⼀3>X-1⼀5+2⼀3,X+1⼀3-(X+2)⼀2>-1⼀6,5(X-A0>3X-A的解为2A<X<-3,求A的取值范围

2025-02-25 02:02:12
推荐回答(1个)
回答1:

X+3/3>X-1/5+2/3
(x+3)/3>(x-1)/5+2/3
[5(x+3)-3(x-1)]/15>2/3
(2x+18)>10
x>-4

X+1/3-(X+2)/2>-1/6
[2(x+1)-3(x+2)]/6>-1/6
(-x-4)>-1
x+4<1
x<-3

5(X-A)>3X-A
5x-5a>3x-a
2x>4a
x>2a

-42a
的交集是:
2a所以
2a>=-4,2a<-3
a>=-2,a<-3/2
-2<=a<-3/2