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lerabond1
@lerabond1
August 2021
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15 задание, пожалуйста)
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treez0r
ОДЗ: x≠0, x>-2, x>-4, x>-14;
x>-2, x≠0
Решение:
log(2) x²+log(2) (x+14)<=log(2) (x+4)+log(2) (x+2)²
log(2) (x²(x+14))/(x+2)²(x+4)<=0
log(2) (x²(x+14))/(x+2)²(x+4)<=log(2) 1
(x²(x+14))/(x+2)²(x+4)<=1
(x²(x+14))-(x+4)(x+2)²)/(x+4)(x+2)²<=0
(x³+14x²-(x+4)(x²+4x+4))/(x+4)(x+2)²<=0
(x³+14x²-x³-4x²-4x-4x²-16x-16)/(x+4)(x+2)²<=0
(6x²-20x-16)/(x+4)(x+2)²<=0
(3x²-15x-8)/(x+4)(x+2)²<=0
(x-((15±(321)^½)/2))/(x+4)(x+2)²<=0
x€(-беск;-4)U[(15-(321)^½)/2;(15+(321)^½)/2]
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Answers & Comments
x>-2, x≠0
Решение:
log(2) x²+log(2) (x+14)<=log(2) (x+4)+log(2) (x+2)²
log(2) (x²(x+14))/(x+2)²(x+4)<=0
log(2) (x²(x+14))/(x+2)²(x+4)<=log(2) 1
(x²(x+14))/(x+2)²(x+4)<=1
(x²(x+14))-(x+4)(x+2)²)/(x+4)(x+2)²<=0
(x³+14x²-(x+4)(x²+4x+4))/(x+4)(x+2)²<=0
(x³+14x²-x³-4x²-4x-4x²-16x-16)/(x+4)(x+2)²<=0
(6x²-20x-16)/(x+4)(x+2)²<=0
(3x²-15x-8)/(x+4)(x+2)²<=0
(x-((15±(321)^½)/2))/(x+4)(x+2)²<=0
x€(-беск;-4)U[(15-(321)^½)/2;(15+(321)^½)/2]