Question

Log0,7 1/(x-2)(x-3)>=log0,7 3/(3-x)(6-x)

Answer 1

Ответ:

х ∈ (-∞; 0] ∪ (3; +∞)

Пошаговое объяснение:

log0,7 1/(x-2)(x-3) ≥ log0,7 3/(3-x)(6-x)

0 < 1/(x-2)(x-3) ≤ 3/(3-x)(6-x)

1/(x-2)(x-3) > 0

(3-x)(6-x) ≤ 3(x-2)(x-3)

(x-2)(x-3) > 0

3(x-2)(x-3) - (3-x)(6-x) ≥ 0

(x-2)(x-3) > 0

(3x-6)(x-3) + (x-3)(6-x) ≥ 0

(x-2)(x-3) > 0

(x-3)2x ≥ 0

х ∈ (-∞; 2) ∪ (3; +∞)

х ∈ (-∞; 0] ∪ [3; +∞)

х ∈ (-∞; 0] ∪ (3; +∞)

Answer 2

To solve the inequality $\log_{0.7}\left(\frac{1}{(x-2)(x-3)}\right) \geq \log_{0.7}\left(\frac{3}{(3-x)(6-x)}\right)$, we can use the property of logarithms that states:

If $\log_a(b) \geq \log_a(c)$, then $b \geq c$, provided that $a$ is a positive number greater than 1.

So, we need to set the expressions inside the logarithms to be greater than or equal to each other:

$\frac{1}{(x-2)(x-3)} \geq \frac{3}{(3-x)(6-x)}$.

To make the algebraic manipulation easier, we can cross-multiply to eliminate the fractions:

$(3-x)(6-x) \geq 3(x-2)(x-3)$.

Next, we expand both sides of the inequality:

$18 - 9x - 6x + x^2 \geq 3(x^2 - 5x + 6)$.

Now, simplify and set the inequality to zero:

$x^2 - 15x + 18 \geq 0$.

To solve the quadratic inequality, we can find the critical points by setting the expression equal to zero:

$x^2 - 15x + 18 = 0$.

Now, factorizing the quadratic equation:

$(x - 3)(x - 6) = 0$.

This gives us two critical points: $x = 3$ and $x = 6$.

We will now create a sign chart to test the inequality in the intervals created by these critical points:

Interval	Test Point	$x^2 - 15x + 18$
$(-\infty, 3)$	$x = 0$	$+$
$(3, 6)$	$x = 4$	$-$
$(6, \infty)$	$x = 7$	$+$

The sign chart indicates that the inequality is satisfied in the intervals $(-\infty, 3)$ and $(6, \infty)$.

Thus, the solution to the original inequality is:

$x \in (-\infty, 3) \cup (6, \infty)$.