①
2x/(x+1)≥1
2x/(x+1)-1≥0
(x-1)/(x+1)≥0
(x-1)(x+1)≥0,且x+1≠0
故x<-1或x≥1
②
(2-x)(x^2-2x-3)≤0
(x-2)(x+1)(x-3)≥0
-1≤x≤2或x≥3
③
(3-x^2)/(x^2+x-2)≥-1
(3-x^2)/(x^2+x-2)+1≥0
(x+1)/(x^2+x-2)≥0
(x+1)/[(x-1)(x+2)]≥0
(x+1)(x-1)(x+2)≥0,且(x-1)(x+2)≠0
所以-2<x≤-1或x>1
2x/(x+1)≥1
解:∵x+1≠0
∴x≠-1
①x+1>0,即x>-1
2x≥x+1
∴x≥1
②x+1<0,即x<-1
2x≤x+1
x≤1
∴x<-1
∴2x/(x+1)≥1的解为:(-∞,-1)U[1,+∞)
(2-x)(x²-2x-3)≤0
解:①2-x=0,即x=2
0≤0
∴x=2
②2-x<0,即x>2时
x²-2x-3≥0
(x-3)(x+1)≥0
x≤-1或x≥3
∴x≥3
③2-x>0,即x<2时
(x-3)(x+1)≤0
-1≤x≤3
∴-1≤x<2
∴(2-x)(x²-2x-3)≤0的解为:[-1,2]U[3,+∞)
(3-x²)/(x²+x-2)≥-1
解: (x²-3)/(x²+x-2)≤-1
x²+x-2≠0
(x+2)(x-1)≠0
∴x≠-2或x≠1
①x²+x-2>0,即x<-2或x>1时
x²-3≤x²+x-2
x≥-1
∴x>1
②x²+x-2<0,即-2
x≤-1
∴-2
三个不等式分别进行化简,可得
(x-1)/(x+1)≥0
(x+1)(x-3)(x-2)≥0
(x+1)/[(x+2)(x-1)]≥0
从是否同号考虑,对应可得
①x≥1或x≤-1 ②x≥3或-1≤x≤2 ③x≥1或-2≤x≤-1
对其取交集可得,x∈{(1,2],[3,+∞)}
擦,我还以为解不等式组呢。
(1)(x-1)(x+1)>=0且(x+1)≠0,即x>=1或x<-1
(2)(2-x)(x^2-2x-3)<=0,即(x-3)(x-2)(x+1)>=0,故而为x>=3或者1=
1.x<=-1或者.x>=1
2.利用穿针引线法
3.方法类似第二问