Question

高一数学不等式求解 谢谢

高一数学不等式求解 谢谢rt

Answer 1

1. 当x>0时,1/x>x=x^2<1, -1<x<1且x>0,  得0<x<1

   当x<0时, -1/x>x,  1/x  + x<0,  (1+x^2)/x<0,  分子>0,  所以x<0

   以上二部分取并,得  x<0或0<x<1

2. x不=1,  当x+1>=0时, ( x+1)-1/(x-1)<0,   (x^2-2)/(x-1)<0,  

   (x-根号2)(x+根号2)(x-1)<0  且x>-1,   x不=1

   x<-根号2 或1<x<根号2    且x>-1,  x不=1

   得:   1<x<根号2

   当x<-1时,  -x-1<1/(x-1),   1/(x-1)+(x+1)>0

   x^2/(x-1)>0,   只要分母x-1>0,     x>1

   但x<-1且x>1  为空集

   所以不等式的解为;    1<x<根号2

3. x/(x+1)<0,   x(x+1)<0, 所以 -1<x<0
   

Answer 2

(1)
|1/x|> x
=> x≠0 (1)

|1/x|> x
1/x^2 > x^2
x^4 < 1
(x^2 -1)(x^2+1) <0
x^2-1<0
-1<x<1 (2)

|1/x|> x
(1) and (2)
x ≠0 and -1<x<1
-1<x<0 or 0<x<1

(2)
|x+1| < 1/(x-1)
=>x>1 (1)

|x+1| < 1/(x-1)
(x+1)^2 . (x-1)^2 <1
(x^2-1)^2 <1
x^4 -2x^2<0
x^2.(x^2 -2) <0
x^2 -2 <0
-√2< x< √2 (2)

|x+1| < 1/(x-1)
(1) and (2)
x>1 and -√2< x< √2
1< x< √2

(3)
|x/(x+1)| > x/(x+1)

=> x≠-1 (1)

|x/(x+1)| > x/(x+1)
=>
x/(x+1) <0
-1<x<0 (2)

|x/(x+1)| > x/(x+1)
(1) and (2)
x≠-1 and -1<x<0
-1<x<0

Answer 3

1. 分x>0和x<02中情况讨论

2. 3，都是分X>-1 和x<-1讨论