23
Представим (х²-х+1)/(х-1)=х+1/(х-1) и (х²-3х+1)/(х-3)=х+1/(х-3)
Получим
2х+1/(х-1)+1/(х-3)>2х-1/(4x+8)
1/(x-1)+1/(x-3)+1/(4x+8)>0
[((x-3)(4x+8)+(x-1)(4x+8)+(x-1)(x-3)]/[(x-1)(x-3)(4x+8)]>0
(4x²-12x+8x-24+4x²-4x+8x-8+x²-3x-x+3)/[(x-1)(x-3)(4x+8)]>0
(9x²-4x-29)/[(x-1)(x-3)(4x+8)]>0
9x²-4x-29=0
D=16+1044=1060
√D=2√265
x1=(4-2√265)/18=(2-√265)/9 U x2=(2+√265)/9
x-1=0⇒x=1
x-3=0⇒x=3
4x+8=0⇒x=-2
_ + _ + _ +
-----------(-2)-------[(2-√265)/9]--------(1)-------[(2-√265)/9]---------(3)-------------
x∈∈(-2;(2-√265)/9] U (1;(2+√265)/9] U (3;∞)
26
Cгруппируем 3/(3+4х)-2/(2+3х) и 1/(1+2х)-4/(4+5х)
3/(3+4х)-2/(2+3х)=(6+9х-6-8х)/(3+4х)(2+3х)=х/(3+4х)(2+3х)
1/(1+2х)-4/(4+5х)=(4+5х-4-8х)/(1+2х)(4+5х)=-3х/(1+2х)(4+5х)
ПОлучили
х/(3+4х)(2+3х)-3х/(1+2х)(4+5х)<0<br>х[(1+2x)(4+5x)-3(3+4x)(2+3x)]/[(3+4x)(2+3x)(1+2x)(4+5x)]<0<br>x(4+5x+8x+10x²-18-27x-24x-36x²)/[(3+4x)(2+3x)(1+2x)(4+5x)]<0<br>x(-26x²-38x-14)/[(3+4x)(2+3x)(1+2x)(4+5x)]<0<br>2x(13x²+19x+7)/[(3+4x)(2+3x)(1+2x)(4+5x)]>0
x=0
13x²+19x+7=0
D=361-364=-3<0⇒13x²+19x+7>0 при любом х
3+4x=0⇒x=-3/4
2+3x=0⇒x=-2/3
1+2x=0⇒x=-1/2
4+5x=0⇒x=-4/5
_ + _ + _ +
-----------(-4/5)-------(-3/4)-------(-2/3)-------(-1/2)----------(0)--------------
x∈∈(-4/5;-3/4) U (-2/3;-1/2) U (0;∞∞)