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AvgustaV
@AvgustaV
October 2021
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log по основанию x+1 числа x^2-x-2≤1
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yarovoe
Logₓ₊₁ x²
-x-2≤1
0<x+1<1, -1<x<0 x
²-x-2>0
-1<x<0
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
+ - - +
---------------------- -1--------------------0---------------2------3------>x
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\
x∈∅
x>0
(x+1)(x-2)>0
x²-2x-3≤0 ,(x+1)(x-3)≤0
////////////////////////////// ////////////////////////////////////////////
--------------------- -1 -------------0---------------2--------------3------------------->x
///////////////////\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈(2;3]
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Answers & Comments
0<x+1<1, -1<x<0 x²-x-2>0
-1<x<0
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
+ - - +
---------------------- -1--------------------0---------------2------3------>x
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\
x∈∅
x>0
(x+1)(x-2)>0
x²-2x-3≤0 ,(x+1)(x-3)≤0
////////////////////////////// ////////////////////////////////////////////
--------------------- -1 -------------0---------------2--------------3------------------->x
///////////////////\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈(2;3]