解不等式(組)6題

2012-02-12 7:51 am
1) -2 < -x < -1

2) -2 > -x > -3

3) (x+1)(x-3) > 0, (x-4)(x-5) > 0

4) (x-3)(x+4) < 0, (x-1)(x+3) > 0

5) (x-1)⁷ (x+6)²¹ (2x-5)³² (3x-2)⁶ (3x+4) (4x²-2x+1) > 0

6) (x-1)⁷ (x+6)²¹ (2x-5)³² (3x-2)⁶ (3x+4) (4x²-2x+1) < 0

回答 (4)

2012-02-14 10:36 pm
✔ 最佳答案
(1) -2 < -x < -1
Multiply by -1, we get 2 > x > 1
Ans (1) : 1 < x < 2
(2) -2 > -x > -3
Mulitply by -1, we get 2 < x < 3
Ans (2) : 2 < x < 3
(3) (x + 1)(x - 3) > 0, (x - 4)(x - 5) > 0
Remember, "Greater" on both sides, "Smaller" in the middle, so, LS is x < -1 or x > 3; and, RS is x < 4 or x >5
Ans (3) : 3 < x < 4 or x > 5
(4) (x - 3)(x + 4) < 0, (x - 1)(x + 3) >0
Same, so, LS is -4 < x < 3; and, RS is x < -3 or x > 1
Ans (4) : -4 < x < -3 or 1 < x < 3
(5) (x - 1)⁷(x + 6)²¹(2x - 5)³²(3x - 2)⁶(3x + 4)(4x² - 2x + 1) > 0
As 4x² - 2x + 1 is always greater than 0, and even power is always greater than or equal to zero, so the inequality equation becomes
(x - 1)(x + 6)(3x + 4) > 0 and x except 5/2, 2/3, so
-6 < x < -4/3 or x > 1 except 5/2, 2/3
Ans (5) : -6 < x < -4/3 or 1 < x < 5/2 or x > 5/2
(6) (x - 1)⁷(x + 6)²¹(2x - 5)³²(3x - 2)⁶(3x + 4)(4x² - 2x + 1) < 0
Same as (5), so the inequality equation becomes
(x - 1)(x + 6)(3x + 4) < 0 and x except 5/2, 2/3, so
x < -6 or -4/3 < x < 1 except 5/2, 2/3
Ans (6) : x < -6 or -4/3 < x < 2/3 or 2/3 < x < 1
2012-02-13 2:06 am
1)-2<-x <-1
1<x<2 2) -2 >-x >-3
2<x<3 3)(x+1)(x-3)>0,(x-4)(x-5)>0
x<-1 or x>3 x<4or x>5
-1>x<3 4>x>5 4)(x-3)(x+4)<0,(x-1)(x+3)>-4<x<3 x<-3 or x>1-4<x<-3 1<x<3

5)[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)>0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)>0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)>0
(x-1)(3x+4)>0,x<>1,x<>-6,3x-2<>0
(x-1)(x+4/3)>0,x<>1,x<>-6,3x-2<>0x<-6 -1<x<-4/3 -4/3<x<1 x>1

6)[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)<0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)<0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)<0
(x-1)(3x+4)<0,x<>1,x<>-6,3x-2<>0
(x-1)(x+4/3)<0,x<>1,x<>-6,3x-2<>0
-4/3<x<1,x<>1,x<>-6,x<>2/3
-4/3<x<-2/3 2/3<x<1
2012-02-12 6:46 pm
To 001:
Your final answer of no.3 should be:
x<-1 or 3<4or x">"5
no.5 and no.6:
D=(-2)^2-4*4*1=-12"<"0
so 4x^2 - 2x + 1 > 0 but not < 0
2012-02-12 9:44 am
1) -2<-x <-1
Sol
1<x<2

2) -2 >-x >-3
Sol
2<x<3

3) (x+1)(x-3)>0,(x-4)(x-5)>0
Sol
(x+1)(x-3)>0=>x<-1 or x>3
(x-4)(x-5)>0=>x<4 or x>5
So
x<-1 or 3<x<4or x<5

4) (x-3)(x+4)<0,(x-1)(x+3)>0
Sol
(x-3)(x+4)<0 =>-4<x<3
(x-1)(x+3)>0 =>x<-3 or x>1
So
-4<x<-3 or 1<x<3

5) [(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)>0
Sol
4x^2-2x+1
D=(-2)^2-4*4*1=-12>0
So
4x^2-2x+1<0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)>0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)>0
(x-1)(3x+4)>0,x<>1,x<>-6,3x-2<>0
(x-1)(x+4/3)>0,x<>1,x<>-6,3x-2<>0x<-6 or -<x<-4/3 or -4/3<x<1 or x>1

6) [(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)<0
Sol
4x^2-2x+1
D=(-2)^2-4*4*1=-12>0
So
4x^2-2x+1>0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)(4x^2-2x+1)<0
[(x-1)^7][(x+6)^21][(2x-5)^32][(3x-2)^6](3x+4)<0
(x-1)(3x+4)<0,x<>1,x<>-6,3x-2<>0
(x-1)(x+4/3)<0,x<>1,x<>-6,3x-2<>0
-4/3<x<1,x<>1,x<>-6,x<>2/3
─4/3<x<2/3or 2/3<x<1




2012-02-12 12:33:19 補充:
4x^2-2x+1<0
改為4x^2-2x+1>0


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