Logₓ₊₁ x²-x-2≤1
0²-x-2>0
-1 x²-x-2≥x+1 решаем сис тему методом интервалов:
D=1²+4·2=9,√D=3, x₁=(1+3)/2=2, x₂=-1, (x+1)(x-2)>0
x²-x-2-x-1≥0, x²-2x-3≥0
D=4+4·3=16, √D=4, x ₁=(2+4)/3=3, x₂=(2-4)/2=-1
(x-3)(x+1)≥0
+ - - +
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x∈∅
x>0
(x+1)(x-2)>0
x²-2x-3≤0 ,(x+1)(x-3)≤0
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x∈(2;3]