Question

((sgrtx)-1/2)^6/((sgrtx)+1/2)^3=1/64

Answer 1

Решение смотрите на картинке.

Answer 2

To solve the given equation:

$\left(\frac{{\sqrt{x} - \frac{1}{2}}}{{\sqrt{x} + \frac{1}{2}}}\right)^6 = \frac{1}{64}$

Let's simplify the equation step by step:

1. Start by multiplying both sides of the equation by $(\sqrt{x} + \frac{1}{2})^3$ to eliminate the denominators:

$\left(\frac{{\sqrt{x} - \frac{1}{2}}}{{\sqrt{x} + \frac{1}{2}}}\right)^6 \cdot (\sqrt{x} + \frac{1}{2})^3 = \frac{1}{64} \cdot (\sqrt{x} + \frac{1}{2})^3$

2. Expand both sides:

$(\sqrt{x} - \frac{1}{2})^6 = \frac{1}{64} \cdot (\sqrt{x} + \frac{1}{2})^3$

3. Simplify the expression on the left side by applying the binomial theorem:

$(\sqrt{x} - \frac{1}{2})^6 = \binom{6}{0} (\sqrt{x})^6 \cdot \left(\frac{-1}{2}\right)^0 + \binom{6}{1} (\sqrt{x})^5 \cdot \left(\frac{-1}{2}\right)^1 + \binom{6}{2} (\sqrt{x})^4 \cdot \left(\frac{-1}{2}\right)^2 + \binom{6}{3} (\sqrt{x})^3 \cdot \left(\frac{-1}{2}\right)^3 + \binom{6}{4} (\sqrt{x})^2 \cdot \left(\frac{-1}{2}\right)^4 + \binom{6}{5} (\sqrt{x})^1 \cdot \left(\frac{-1}{2}\right)^5 + \binom{6}{6} (\sqrt{x})^0 \cdot \left(\frac{-1}{2}\right)^6$