Question

5^(1/x-1)+5^(1/x+2)больше либо равно 126

Answer 1

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Answer 2

To solve the inequality $5^{\frac{1}{x}-1} + 5^{\frac{1}{x}+2} \geq 126$, we'll first simplify it and then find the appropriate values of $x$.

Step 1: Combine the terms with the same base (5) by using the properties of exponents. Recall that $a^m \cdot a^n = a^{m+n}$.

$5^{\frac{1}{x}-1} + 5^{\frac{1}{x}+2} = 5^{\frac{1}{x}-1} + 5^{\frac{1}{x}} \cdot 5^{2}$

Step 2: Convert all terms to a common base, if possible. In this case, we can express $5^2$ as $25$ since both represent the same value.

$5^{\frac{1}{x}-1} + 5^{\frac{1}{x}} \cdot 25$

Step 3: Now, we'll try to eliminate the fractional exponent. Recall that $a^{-m} = \frac{1}{a^m}$. So, $5^{\frac{1}{x}-1} = \frac{1}{5^{1-\frac{1}{x}}} = \frac{1}{5^{\frac{x-1}{x}}}$.

Replace the term $5^{\frac{1}{x}-1}$ in the original inequality:

$\frac{1}{5^{\frac{x-1}{x}}} + 5^{\frac{1}{x}} \cdot 25 \geq 126$

Step 4: Multiply both sides by $5^{\frac{x}{x}}$ to get rid of the fractions in the exponents.

$5^{\frac{x}{x}} \cdot \frac{1}{5^{\frac{x-1}{x}}} + 5^{\frac{x}{x}} \cdot 5^{\frac{1}{x}} \cdot 25 \geq 126 \cdot 5^{\frac{x}{x}}$

Step 5: Simplify further.

$5^{1-\frac{x-1}{x}} + 5^{\frac{x}{x}} \cdot 25 \cdot 5^{\frac{1}{x}} \geq 126 \cdot 5^{\frac{x}{x}}$

Step 6: More simplification.

$5^{\frac{1}{x}} + 25 \cdot 5^{\frac{1}{x}} \geq 126 \cdot 5^{\frac{x}{x}}$

Step 7: Finally, simplify $5^{\frac{x}{x}}$