Question

13-5•(3^х) ————————— ≥0,5 (9^х)-12•(3^х)+27

Answer 1

Ответ:

13–5·3x / (9x–12·3x+27) ≥ 0.5⪻⪼t^2-2t+1/t^2-12t+27≤0⪻⪼(t-1)^2/(t-3)(t-9)≤0 t=1 1<x<2

Answer 2

$\frac{13-5\cdot 3^x}{9^x-12\cdot 3^x+27}\geq 0,5\Leftrightarrow \frac{-5\left ( 3^x-3 \right )}{3^x\left ( 3^x-3 \right )-9\left ( 3^x-3 \right )}\geq 0,5\Leftrightarrow -\frac{5}{3^x-9}\geq 0,5\\\frac{1+3^x}{3^x-9}\leq 0\Rightarrow 3^x\in \left [ -1;9 \right )\Rightarrow x\in \left ( -\infty ;2 \right )\setminus \left \{ 1 \right \}$

Answer 3

To solve the inequality:

13 - 5*(3^x) / (9^x) - 12*(3^x) + 27 ≥ 0.5

Let's simplify the expression first:

13 - 5*(3^x) / (3^2x) - 12*(3^x) + 27 ≥ 0.5

13 - 5*(3^x) / (3^2x) - 12*(3^x) + 27 - 0.5 ≥ 0

13 - 5*(3^x) / (3^2x) - 12*(3^x) + 26.5 ≥ 0

To solve this inequality, we can make a substitution. Let's substitute y = 3^x:

13 - 5y / (9y) - 12y + 26.5 ≥ 0

Now, let's simplify further:

13 - 5/9 - 12y + 26.5 ≥ 0

28.5 - 12y ≥ 0

-12y ≥ -28.5

Dividing by -12, remember to flip the inequality:

y ≤ 28.5 / 12

y ≤ 2.375

Since y = 3^x, we can substitute back:

3^x ≤ 2.375

To solve for x, we take the logarithm of both sides (with base 3):

x ≤ log3(2.375)

Using a calculator, we find:

x ≤ 0.4503 (approximately)

Therefore, the solution to the inequality is:

x ≤ 0.4503