Question

Даю 100 баллов алгебра​

Answer 1

Объяснение:

A)

log(2)(2-5x)>1 одз : х<2/5

2-5х>2¹

2-5х>2

-5х>2-2

-5х>0

х< 0

х∈(-∞;0)

Б)

log(0,2)(4-2x)> -1 одз: х<2

4-2х<0,2`¹

4-2х<(1/5)`¹

4-2х<5

-2х<5-4

-2х<1

х> -1/2 ; х<2

х∈( -1/2;2)

В)

log(0,4)(2x-5)>log(0,4)(x+1) одз: (5/2;+∞)

2х-5<х+1

2х-х<1+5

х<6 х∈(5/2;+∞)

х∈(5/2;6)

Г)

log(0,3)(x-3 / 1-x)≤0 одз: (1;3)

х-3 / 1-х ≥0,3⁰

х-3 / 1-х ≥ 1

х-3 / 1-х -1≥0

(х-3-1+х) / 1-х ≥ 0

2х-4 / 1-х ≥ 0

2(х-2) / 1-х ≥0

2(х-2)≥0 2(х-2)≤0

1-х>0 1-х<0

х≥0 х≤2

х<1 х>1

ø х∈(1;2]

х∈(1;2] х∈(1;3)

х∈(1;2]

Answer 2

$\displaystyle\bf\\1)\\\\ODZ \ : \ 2-5x > 0 \ \ \Rightarrow \ \ \ -5x > -2 \ \ \Rightarrow \ \ \boxed{x < 0,4}\\\\\log_{2} (2-5x) > 1\\\\2 > 1 \ \ \Rightarrow \ \ 2-5x > 2\\\\-5x > 0\\\\x < 0\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ 0\Big)\\\\\\2)\\\\ODZ \ : \ 4-2x > 0 \ \ \Rightarrow \ \ \ -2x > -4 \ \ \Rightarrow \ \ \boxed{x < 2}\\\\\log_{0,2} (4-2x) > -1\\\\0 < 0,2 < 1 \ \ \Rightarrow \ \ 4-2x < 5\\\\-2x < 1\\\\x > -0,5\\\\Otvet \ : \ x\in\Big(-0,5 \ ; \ 2\Big)$

$\displaystyle\bf\\3)\\\\ODZ \ : \ \left \{ {{2x-5 > 0} \atop {x+1 > 0}} \right. \ \ \Rightarrow \ \ \left \{ {{x > 2,5} \atop {x > -1}} \right. \ \ \Rightarrow \ \ x > 2,5\\\\\\\log_{0,4} (2x-5) > \log_{0,4}(x+1)\\\\0 < 0,4 < 1 \ \ \Rightarrow \ \ 2x-5 < x+1\\\\2x-x < 1+5\\\\x < 6\\\\Otvet \ : \ x\in\Big(2,5 \ ; \ 6\Big)\\\\\\4)\\\\ODZ \ : \ \frac{x-3}{1-x} > 0\\\\\\\frac{x-3}{x-1} < 0\\\\\\+ + + + + (1) - - - - - (3) + + + + + \\\\x\in\Big(1 \ ; \ 3\Big)\\\\\\\log_{0,3} \frac{x-3}{1-x} \leq 0$

$\displaystyle\bf\\0 < 0,3 < 1 \ \ \Rightarrow \ \ \frac{x-3}{1-x} \geq 1\\\\\\\frac{x-3}{1-x} -1\geq 0\\\\\\\frac{x-3-1+x}{1-x} \geq 0\\\\\\\frac{2x-4}{1-x} \geq 0\\\\\\\frac{2(x-2)}{x-1} \leq 0 \ \ , \ \ x\neq 1\\\\\\+ + + + + (1) - - - - - [2] + + + + + \\\\\\x\in \Big(1 \ ; \ 2\Big]\\\\\\Otvet \ : \ x\in\Big(1 \ ; \ 2\Big]$