Question

1) (1/2)^-5 > (1/2)^2х+3 2) 5^х^2 <=25 решите пажалуйста

Answer 1

Ответ:

1) $(-4;\infty)$ 2) $[-\sqrt{2}; \sqrt{2} ]$

Объяснение:

1)

$(\frac{1}{2})^{-5}>(\frac{1}{2} )^{2x+3}\\\\ -5-8\\x>-4$

2)

$5^{x^2}\leq 25\\5^{x^2}\leq 5^2\\x^{2}\leq 2\\(x-\sqrt{2})(x+\sqrt{2})\leq 0\\ -\sqrt{2}\leq x\leq \sqrt{2}$

Answer 2

1. To solve the inequality (1/2)^(-5) > (1/2)^(2x+3), we can compare the exponents and simplify the inequality:

(1/2)^(-5) > (1/2)^(2x+3)

Using the rule of exponents that states (a^m)/(a^n) = a^(m-n), we can rewrite the inequality as:

2^5 > 2^(2x+3)

Simplifying further:

32 > 2^(2x+3)

Now, we can compare the exponents:

2x + 3 < 5

Subtracting 3 from both sides:

2x < 2

Dividing by 2:

x < 1

Therefore, the solution to the inequality is x < 1.

2. To solve the inequality 5^(x^2) <= 25, we can take the square root of both sides to eliminate the exponent:

√(5^(x^2)) <= √25

Simplifying:

5^(x^2/2) <= 5

Since the base is the same (5), we can equate the exponents:

x^2/2 <= 1

Multiplying both sides by 2:

x^2 <= 2

Taking the square root of both sides (considering both positive and negative roots):

x <= √2 or x >= -√2

Therefore, the solution to the inequality is x <= √2 or x >= -√2.