Помогите с логарифмическим неравенством:
[tex]log_{2} [/tex](2x-1)-2≥[tex]log_{2} [/tex](x+2)-[tex]log_{2} [/tex](x+1)
[tex]log_{2} [/tex](2x-1)-2≥[tex]log_{2} [/tex](x+2)-[tex]log_{2} [/tex](x+1)
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Verified answer
Log₂(2x-1)-2≥log₂(x+2)-log₂(x+1)D(y): 2x-1>0, x+2>0, x+1>0
2x>1, x>-2, x>-1
x>0,5
log₂(2x-1)-log₂2²≥log₂(x+2)/(x+1)
log₂(2x-1)/4≥log₂(x+2)/(x+1)
(2x-1)/4≥(x+2)/(x+1) /*4(x+1)
(2x-1)(x+1)≥4(x+2)
2x²+2x-x-1≥4x+8
2x²-3x-9≥0
D=(-3)²+4*2*9=9+72=81
x=(3+9)/2*2=3
x=(3-9)/2*2=-1,5
2(x-3)(x+1,5)≥0
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-1,5 0,5 3
Ответ: x∈[3;+∞)