решение задания смотри на фотографии
(x-3+x-2+x-1)/(x-1)(x-2)(x-3)≤1
(3x-6)/(x-1)(x-2)(x-3)≤1
3(x-2)/(x-1)(x-2)(x-3)≤1
3/(x-1)(x-3) -1≤0
(3-x²+4x-3)/(x-1)(x-3)≤0
(4x-x²)/(x-1)(x-3)≤0
x(4-x)/(x-1)(x-3)≤0
x=0 x=4 x=1 x=3
_ + _ + _
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0 1 3 4
x∈(-∞;0] U (1;3) U [4;∞)
Построила в Екселе с таблицей, взяла х(-10;10) с щагом 1.
ОДЗ : x > 0
+ - +
__________₀__________₀__________
0 2
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_______________₀__________________
1
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x ∈ (2 ; + ∞)
+ - +
_______________₀_____________₀_________
0 2
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_______________________₀_______________
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 1
x ∈ (0 ; 1)
Ответ : (0 ; 1) ∪ (2 , + ∞)