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deanonm
@deanonm
July 2022
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log √2 (x^2+10x)>= log √2 (x-14)
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m11m
Verified answer
ОДЗ:
1) x² +10x>0
x(x+10)>0
x=0 x= -10
+ - +
-
----------
-10 ------------- 0
------------
\\\\\\\\\\\\\ \\\\\\\\\\\\\
x∈(-∞; -10)U(0; +∞)
2) x-14>0
x>14
В итоге ОДЗ: х∈(14; +∞)
Так как основание логарифма √2>0, то
x²+10x≥x-14
x²+10x-x+14≥0
x²+9x+14≥0
x²+9x+14=0
D=9² -4*14=81-56=25
x₁=(-9-5)/2= -7
x₂=(-9+5)/2= -2
+ - +
-
---------- -7
--------------
-2 -------------
\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\
x∈(-∞; -7]U[-2; +∞)
С учетом ОДЗ получаем
х∈(14; +∞)
Ответ: (14; +∞)
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Answers & Comments
Verified answer
ОДЗ:1) x² +10x>0
x(x+10)>0
x=0 x= -10
+ - +
----------- -10 ------------- 0 ------------
\\\\\\\\\\\\\ \\\\\\\\\\\\\
x∈(-∞; -10)U(0; +∞)
2) x-14>0
x>14
В итоге ОДЗ: х∈(14; +∞)
Так как основание логарифма √2>0, то
x²+10x≥x-14
x²+10x-x+14≥0
x²+9x+14≥0
x²+9x+14=0
D=9² -4*14=81-56=25
x₁=(-9-5)/2= -7
x₂=(-9+5)/2= -2
+ - +
----------- -7 -------------- -2 -------------
\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\
x∈(-∞; -7]U[-2; +∞)
С учетом ОДЗ получаем
х∈(14; +∞)
Ответ: (14; +∞)