I x^2-11x+30 | ←делить на 1-x = x^2 -12 x+ 36
вероятно так...
I x^2-11x+30 |/(1-x)=x^2 -12 x+ 36
I(x-6)(x-5)I/(1-x)=(x-6)²
+ -
----1-----------------------------------
+ - +
---------------5-----------------6-------------
x∈(-∞;5)∪[6;+∞)
(x-6)(x-5)/(1-x)=(x-6)² (x-6)(x-5)-(1-x)·(x-6)² =0 (x-6)[(x-5)+(x-1)(x-6)]=0
(x-6)(x²-6x+1)=0
x1=6
x2=3-√8 ∉(-∞;5]∪[6;+∞)
x3=3+√8
x∈[5;6)
-(x-6)(x-5)/(1-x)=(x-6)² (x-6)(x-5)=(x-6)²(x-1) (x-6)[(x-5)-(x-6)(x-1) ]=0
(x-6)(-x²+8x-11)=0
(x-6)(x²-8x+11)=0
x1=6
x2=4-√5∉[5;6)
x3=4+√5∉[5;6)
ответ:
x1=6 x2=3+√8