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
(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
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
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
1) －2<－x <－1
Sol
1<x<2
2) －2 >－x >－3
2<x<3
3) (x+1)(x－3)>0，(x－4)(x－5)>0
(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
(x－3)(x+4)<0 =>－4<x<3
(x－1)(x+3)>0 =>x<－3 or x>1
－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
4x^2－2x+1
D=(－2)^2－4*4*1=－12>0
4x^2－2x+1<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
4x^2－2x+1>0
─4/3<x<2/3or 2/3<x<1
2012-02-12 12:33:19 補充：
改為4x^2－2x+1>0
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Answers & Comments
Verified answer
(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
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
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
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