1)x<-1 U x>1 -4(x²-1)-3≥1/(x²-1) (-4(x²-1)²-3(x²-1)-1)/(x²-1)≥0 (4(x²-1)²+3(x²-1)+1)/(x²-1)≤0 x²-1=a (4a²+3a+1)/a≥0 4a²+3a+1>0 при любом а,т.к D<0⇒a<0 x²-1<0⇒-1<x<1 не удов усл нет решения 2)-1<x<1 4(x²-1)-3≥1/(x²-1) (4(x²-1)²-3(x²-1)-1)/(x²-1)≥0 x²-1=a (4a²-3a-1)/a≥0 4a²-3a-1=0 D=9+16=25 a1=(3-5)/8=-1/4 U a2=(3+5)/8=1 a=0 _ + _ + [-1/4](0)[1] -1/4≤a<0 U a≥1 {x²-1≥-1/4⇒x²-3/4≥0⇒x≤-√3/2 U x≥√3/2 {x²-1<0⇒-1<x<1 -1<x≤-√3/2 U √3/2≤x<1 x²-1≥1⇒x²-2≥0⇒x≤-√2 U x≥√2 ответ x∈(-1;-√3/2] U [√3/2;1)
-4(x²-1)-3≥1/(x²-1)
(-4(x²-1)²-3(x²-1)-1)/(x²-1)≥0
(4(x²-1)²+3(x²-1)+1)/(x²-1)≤0
x²-1=a
(4a²+3a+1)/a≥0
4a²+3a+1>0 при любом а,т.к D<0⇒a<0
x²-1<0⇒-1<x<1 не удов усл
нет решения
2)-1<x<1
4(x²-1)-3≥1/(x²-1)
(4(x²-1)²-3(x²-1)-1)/(x²-1)≥0
x²-1=a
(4a²-3a-1)/a≥0
4a²-3a-1=0
D=9+16=25
a1=(3-5)/8=-1/4 U a2=(3+5)/8=1
a=0
_ + _ +
[-1/4](0)[1]
-1/4≤a<0 U a≥1
{x²-1≥-1/4⇒x²-3/4≥0⇒x≤-√3/2 U x≥√3/2
{x²-1<0⇒-1<x<1
-1<x≤-√3/2 U √3/2≤x<1
x²-1≥1⇒x²-2≥0⇒x≤-√2 U x≥√2
ответ x∈(-1;-√3/2] U [√3/2;1)