Question

Ix² - 2x+1I I x-1I
----------- + ---------- - 12< 0
Ix²-4x+4 I I x-2I

Odp: x∈(-∞, 1,75) u (2,5 , +∞)

Dam najlepszą jeżeli ktoś mi to rozwiążę :)

Answer 1

Ix² - 2x+1I I x-1I
----------- + ---------- - 12< 0 x≠2
Ix²-4x+4 I I x-2I

|(x-1)²| I x-1I
________ + _____ - 12 < 0
|(x-2)²| I x-2I

|(x-1)²| + I(x-1)(x-2)I - 12|(x-2)²|
_______________________________ < 0
|(x-2)²|

(x-1)² - 12(x-2)² + I(x-1)(x-2)I < 0
x²-2x+1 - 12(x²-4x+4) + I(x-1)(x-2)I < 0
x²-2x+1-12x²+48x-48 + I(x-1)(x-2)I < 0
|x-1|*|x-2| < 11x²-46x+47

dla x∈(-∞,1)
-(x-1)*(-(x-2)) < 11x²-46x+47
(x-1)*(x-2) < 11x²-46x+47
x²-3x+2<11x²-46x+47
10x²-43x+45>0
(2x-5)(5x-9)>0
x∈(-∞,1⅘)U(2½,∞)
zatem x∈(-∞,1)

dla x∈<1,2)
-(x-2)(x-1) < 11x²-46x+47
-x²+3x-2<11x²-46x+47
12x²-49x+49>0
(3x-7)(4x-7)>0
x∈(-∞,1¾)U(2⅓; ∞)
zatem x∈<1,1¾)

dla x∈<2,∞)
(x-1)*(x-2) < 11x²-46x+47
x²-3x+2<11x²-46x+47
10x²-43x+45>0
(2x-5)(5x-9)>0
x∈(-∞,1⅘)U(2½,∞)
zatem x∈(2½,∞)

odp. x∈(-∞,1¾)U(2½,∞)