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НоВыЙ13
@НоВыЙ13
July 2022
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Решите уравнение:
[tex]2log_{2}(1- \frac{13}{2x+7} )=3log_{2}(2+ \frac{13}{x-3} )+2[/tex]
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contrlc
Verified answer
Ответ::::::::::::::::::::
3 votes
Thanks 1
sedinalana
Verified answer
1-13/(2x+7)=(2x+7-13)/(2x+7)=(2x-6)/(2x+7)=2(x-3)/(2x+7)
2+13/(x-3)=(2x-6+13)/(x-3)=(2x+7)/(x-3)
--------------------------------
2log(2)[2(x-3)/(2x+7)]=3log(2)[(2x+7)/(x-3)]+2
ОДз
(x-3)/(2x+7)>0
x=3 x=-3,5
x∈(-∞;-3,5) U (3;∞)
log(2)[2(x-3)/(2x+7)]²=log(2)[4(2x+7)³/(x-3)]
4(x-3)²/(2x+7)²=4(2x+7)³/(x-3)³
(x-3)²/(2x+7)²=(2x+7)³/(x-3)³
(x-3)^5=(2x+7)^5
x-3=2x+7
2x-x=-3-7
x=-10
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Answers & Comments
Verified answer
Ответ::::::::::::::::::::Verified answer
1-13/(2x+7)=(2x+7-13)/(2x+7)=(2x-6)/(2x+7)=2(x-3)/(2x+7)2+13/(x-3)=(2x-6+13)/(x-3)=(2x+7)/(x-3)
--------------------------------
2log(2)[2(x-3)/(2x+7)]=3log(2)[(2x+7)/(x-3)]+2
ОДз
(x-3)/(2x+7)>0
x=3 x=-3,5
x∈(-∞;-3,5) U (3;∞)
log(2)[2(x-3)/(2x+7)]²=log(2)[4(2x+7)³/(x-3)]
4(x-3)²/(2x+7)²=4(2x+7)³/(x-3)³
(x-3)²/(2x+7)²=(2x+7)³/(x-3)³
(x-3)^5=(2x+7)^5
x-3=2x+7
2x-x=-3-7
x=-10