Question

Прошу помогите!!!!Даю 40 балов!!!!
$log3(x + 3) - log3(x - 1) > 1 - log3(4 - x)$
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Answer 1

Ответ:  x∈(1,3).

Объяснение:

$log_3(x+3)-log_3(x-1) > 1-log_3(4-x)$

ОДЗ:

$x+3 > 0\ \ \ \ \ \ x > -3\\\\x-1 > 0\ \ \ \ \ \ x > 1\\\\4-x > 0\ \ \ \ \ \ x < 4\ \ \ \ \ \ \Rightarrow\\\\x\in(1;4).$

$log_3(x+3)-log_3(x-1) > log_33-log_3(4-x)\\\displaystyle\\log_3\frac{x+3}{x-1} > log_3\frac{3}{4-x} \ \ \ \ \ \ \Rightarrow\\\\\frac{x+3}{x-1} > \frac{3}{4-x} \\\\\frac{x+3}{x-1} -\frac{3}{4-x} > 0\\\\\frac{(x+3)*(4-x)-3*(x-1)}{(x-1)*(4-x)} > 0\\\\\frac{4x+12-x^2-3x-3x+3}{(x-1)*(4-x)} > 0\\\\\frac{-x^2-2x+15}{(x-1)*(4-x)} > 0\\\\\frac{-(x^2+2x-15)}{(x-1)*(4-x)} > 0\\\\\frac{x^2+2x-15}{(x-1)*(x-4)} > 0$

$\displaystyle\\\frac{x^2+5x-3x-15}{(x-1)*(x-4)} > 0\\\\\frac{x*(x+5)-3*(x+5)}{(x-1)*(x-4)} > 0\\\\\frac{(x+5)*(x-3)}{(x-1)*(x-4)} > 0$

-∞__+__-5__-__1__+__3__-__4__+__+∞           ⇒

x∈(-∞;-5)U(1;3)U(4;+∞).

Учитывая ОДЗ: x∈(1;3).