прошу хелп !!!!!!!!!!!!!!!!!!!!!!. Created by Miа16. algebra-ru

1) ОДЗ : 3x - 5 > 03x > 5x > 1 2/3ОДЗ :x² - 2x > 0x(x - 2) > 0x ∈ (- ∞ ; 0) ∪ (2 ; + ∞)3) ОДЗ :

прошу хелп !!!!!!!!!!!!!!!!!!!!!!...

Universalka

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1) ОДЗ :

3x - 5 > 0

3x > 5

x > 1 2/3

$lg(3x-5)=0\\\\3x-5=10^{0}\\\\3x-5=1\\\\3x=6\\\\x=2\\\\Otvet:\boxed{2}$

ОДЗ :

x² - 2x > 0

x(x - 2) > 0

x ∈ (- ∞ ; 0) ∪ (2 ; + ∞)

$log_{2}(x^{2}-2x)=3\\\\x^{2}-2x=2^{3}\\\\x^{2}-2x=8\\\\x^{2}-2x-8=0\\\\x_{1} =-2\\\\x_{2} =4\\\\Otvet:\boxed{-2;4}$

3) ОДЗ :

$\left \{ {{x^{2}-3x+2>0 } \atop {2x-4>0}} \right. \\\\\left \{ {{(x-1)(x-2)>0} \atop {x>2}} \right.\\\\\left \{ {{x\in(-\infty;1)\cup(2;+\infty)} \atop {x>2}} \right.\Rightarrow x\in(2;+\infty)\\\\\\log_{2}(x^{2}-3x+2)=log_{2}(2x-4)\\\\x^{2}-3x+2=2x-4\\\\x^{2}-5x+6=0\\\\x_{1}=3\\\\x_{2}=2-neyd\\\\Otvet:\boxed{3}$

2 votes Thanks 2

sharofat0

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Ответ:

1) х=2

2) х_1=-2

х_2=4

3) х=3

Решить логарифмическое

уравнение.

2 votes Thanks 1

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