Question

1/9)^x-6(1/3)^x>-9 решите

Answer 1

Ответ:

ушарко цзаьстеиеок

Объяснение:

епаровадьтимоха

Answer 2

To solve the inequality $\left(\frac{1}{9}\right)^x - 6\left(\frac{1}{3}\right)^x > -9$, we can use a substitution. Let's set $y = \left(\frac{1}{3}\right)^x$. Now the inequality becomes:

$\left(\frac{1}{9}\right)^x - 6y > -9$

Next, rewrite the expression on the left-hand side in terms of $y$:

$\left(\frac{1}{9}\right)^x = \left(\left(\frac{1}{3}\right)^x\right)^2 = y^2$

Now the inequality becomes:

$y^2 - 6y > -9$

Move all terms to the left-hand side:

$y^2 - 6y + 9 > 0$

Now, we need to find the values of $y$ (or $\left(\frac{1}{3}\right)^x$) for which this quadratic expression is greater than zero.

Factorize the quadratic expression:

$(y - 3)^2 > 0$

Since $(y - 3)^2$ is always non-negative (equal to zero at $y = 3$), we know that it will always be greater than zero except at $y = 3$. Thus, the inequality holds for all values of $y$ except $y = 3$.

Now, recall that $y = \left(\frac{1}{3}\right)^x$. So the inequality holds for all values of $\left(\frac{1}{3}\right)^x$ except $\left(\frac{1}{3}\right)^x = 3$.

Now, solve for $x$:

$\left(\frac{1}{3}\right)^x = 3$

Taking the logarithm of both sides (any base will work, but we'll use the natural logarithm):

$x \cdot \ln\left(\frac{1}{3}\right) = \ln(3)$

$\ln\left(\frac{1}{3}\right)$ is negative, so divide both sides by it (which is equivalent to multiplying by its reciprocal):

$x = \frac{\ln(3)}{\ln\left(\frac{1}{3}\right)}$

Using a calculator to evaluate this, we get:

$x \approx -2.7095$

So, the solution to the inequality is $x < -2.7095$ or $x > 2.7095$ (approximately).