Răspuns :
a) cazul I
-1≤(x²+x)/(x²-6x+8)
-(x²-6x+8)≤x²+x
-x²+6x-8-x²-x≤0
-2x²+5x-8≤0 inmultim -1
2x²-5x+8≥0
2x²-5x+8=0
Δ=25-64=-39
Δ<0
cazul 2
(x²+x)/(x²-6x+8)≤3
x²+x≤3(x²-6x+8)
x²+x≤3x²-18x+24
x²-3x²+x+18x-24≤0
-2x²+19x-24≤0 inmultim -1
2x²-19x+24≥0
Δ=361-192=169
x₁=(19-13)/4=6/4=3/2
x₂=(19+13)/4=32/4=8
x -∞ 3/2 8 +∞
f(x) + + + + + + 0- - - - - - - - 0++++++++++++
x∈(-∞,3/2]U[8,+∞)
c) cazul 1
(x²-3x+7)/(x²-9)≤1
x²-3x+7≤x²-9
x²-x²-3x+7+9≤0
-3x≤-16 inbmultim -1
3x≥16
x≥16/3
x∈[16/3,+∞)
cazul 2
-(x²-3x+7)/(x²-9)≤1 inmultim -1
(x²-3x+7)/(x²-9)≤-1
x²-3x+7≤-x²+9
x²+x²-3x+7-9≤0
2x²-3x-2≤0
Δ=9+16=25
x1=(3-5)/4=-2/4=-1/2
x2=(3+5)/4=8/4=2
x∈[-1/2,2]
-∞ -1/2 2 16/3 +∞
[//////////////////////////
[/////////////////////]
deci x∈[-1/2]U[16/3,+α)
b) cazul 1
-1≤(2x²-x+3)/(x²+x-6)
-x²-x+6≤2x²-x+3
-x²-2x²-x+x+6-3≤0
-3x²+3≤0 :(-3)
x²-1≥0
(x+1)(x-1)≥0
x1=-1
x2=1
x∈(-∞,-1]U[1,+∞)
cazul 2
(2x²-x+3)/(x²+x-6)≤2
2x²-x+3≤2(x²+x-6)
2x²-x+3≤2x²+2x-12
2x²-2x²-x-2x≤-12-3
-3x≤-15 inmultim -1
3x≥15
x≥15/3
x≥5
x∈[5,+∞)
-∞ -1 1 5 +∞
////////////////////] [//////////////////////////////////////////
[//////////////////////////
deci x∈[5,+∞)
d) cazul 1
(x²+x+1)/(4x²-1)>1
x²+x+1>4x²-1
x²-4x²+x+1+1>0
-3x²+x+2>0 inmultim -1
3x²-x-2<0
3x²-x-2=0
Δ=1+24=25
x1=(1-5)/6=-4/6=-2/3
x2=(1+5)/6=6/6=1
x -∞ -2/3 1 +∞
f(x) +++++++++0- - - - - - - - 0+++++++++++
x∈(-2/3,1)
cazul 2
-(x²+x+1)/((4x²-1)>1 inmultim -1
(x²+x+1)/(4x²-1)<-1
x²+x+1<-4x²+1
x²+4x²+x+1-1<0
5x²+x<0
x(5x+1)<0
x1=0
5x+1=0
5x=-1
x=-1/5
x -∞ -1 -1/5 +∞
f(x) ++++++++0 - - - - - - - 0++++++++++++
x∈(-1, -1/5)
-∞ -1 -2/3 -1/5 1
(/////////////////////////////////////////////)
(///////////////////////////////////)
deci x∈(-2/3,-1/5)
e) cazul 1
(3x²-8x+6)/(-x²+5x-4)<1
3x²-8x+6<-x²+5x-4
3x²+x²-8x-5x+6+4<0
4x²-13x+10<0
Δ=169-160=9
x1=(13-3)/8=10/8=5/4
x2=(13+3)/5=16/8=2
x -∞ 5/4 2 +∞
f(x) +++++++++++0- - - - - - - - - 0++++++++++++++
x∈(5/4,2)
caZUL 2
-(3x²-8x+6)/(-x²+5x-4)<1 inmultim -1
(3x²-8x+6)/(-x²+5x-4)<-1
3x²-8x+6<x²-5x+4
3x²-x²-8x+5x+6-4<0
2x²-3x+2<0
Δ=9-16=-7
Δ<0
x∈(5/4,2)
-1≤(x²+x)/(x²-6x+8)
-(x²-6x+8)≤x²+x
-x²+6x-8-x²-x≤0
-2x²+5x-8≤0 inmultim -1
2x²-5x+8≥0
2x²-5x+8=0
Δ=25-64=-39
Δ<0
cazul 2
(x²+x)/(x²-6x+8)≤3
x²+x≤3(x²-6x+8)
x²+x≤3x²-18x+24
x²-3x²+x+18x-24≤0
-2x²+19x-24≤0 inmultim -1
2x²-19x+24≥0
Δ=361-192=169
x₁=(19-13)/4=6/4=3/2
x₂=(19+13)/4=32/4=8
x -∞ 3/2 8 +∞
f(x) + + + + + + 0- - - - - - - - 0++++++++++++
x∈(-∞,3/2]U[8,+∞)
c) cazul 1
(x²-3x+7)/(x²-9)≤1
x²-3x+7≤x²-9
x²-x²-3x+7+9≤0
-3x≤-16 inbmultim -1
3x≥16
x≥16/3
x∈[16/3,+∞)
cazul 2
-(x²-3x+7)/(x²-9)≤1 inmultim -1
(x²-3x+7)/(x²-9)≤-1
x²-3x+7≤-x²+9
x²+x²-3x+7-9≤0
2x²-3x-2≤0
Δ=9+16=25
x1=(3-5)/4=-2/4=-1/2
x2=(3+5)/4=8/4=2
x∈[-1/2,2]
-∞ -1/2 2 16/3 +∞
[//////////////////////////
[/////////////////////]
deci x∈[-1/2]U[16/3,+α)
b) cazul 1
-1≤(2x²-x+3)/(x²+x-6)
-x²-x+6≤2x²-x+3
-x²-2x²-x+x+6-3≤0
-3x²+3≤0 :(-3)
x²-1≥0
(x+1)(x-1)≥0
x1=-1
x2=1
x∈(-∞,-1]U[1,+∞)
cazul 2
(2x²-x+3)/(x²+x-6)≤2
2x²-x+3≤2(x²+x-6)
2x²-x+3≤2x²+2x-12
2x²-2x²-x-2x≤-12-3
-3x≤-15 inmultim -1
3x≥15
x≥15/3
x≥5
x∈[5,+∞)
-∞ -1 1 5 +∞
////////////////////] [//////////////////////////////////////////
[//////////////////////////
deci x∈[5,+∞)
d) cazul 1
(x²+x+1)/(4x²-1)>1
x²+x+1>4x²-1
x²-4x²+x+1+1>0
-3x²+x+2>0 inmultim -1
3x²-x-2<0
3x²-x-2=0
Δ=1+24=25
x1=(1-5)/6=-4/6=-2/3
x2=(1+5)/6=6/6=1
x -∞ -2/3 1 +∞
f(x) +++++++++0- - - - - - - - 0+++++++++++
x∈(-2/3,1)
cazul 2
-(x²+x+1)/((4x²-1)>1 inmultim -1
(x²+x+1)/(4x²-1)<-1
x²+x+1<-4x²+1
x²+4x²+x+1-1<0
5x²+x<0
x(5x+1)<0
x1=0
5x+1=0
5x=-1
x=-1/5
x -∞ -1 -1/5 +∞
f(x) ++++++++0 - - - - - - - 0++++++++++++
x∈(-1, -1/5)
-∞ -1 -2/3 -1/5 1
(/////////////////////////////////////////////)
(///////////////////////////////////)
deci x∈(-2/3,-1/5)
e) cazul 1
(3x²-8x+6)/(-x²+5x-4)<1
3x²-8x+6<-x²+5x-4
3x²+x²-8x-5x+6+4<0
4x²-13x+10<0
Δ=169-160=9
x1=(13-3)/8=10/8=5/4
x2=(13+3)/5=16/8=2
x -∞ 5/4 2 +∞
f(x) +++++++++++0- - - - - - - - - 0++++++++++++++
x∈(5/4,2)
caZUL 2
-(3x²-8x+6)/(-x²+5x-4)<1 inmultim -1
(3x²-8x+6)/(-x²+5x-4)<-1
3x²-8x+6<x²-5x+4
3x²-x²-8x+5x+6-4<0
2x²-3x+2<0
Δ=9-16=-7
Δ<0
x∈(5/4,2)