Представьте в виде рациональной дроби:
а) $\frac{x - \frac{yz}{y - z}}{y - \frac{xz}{x - z}}$;
б) $\frac{\frac{a - x}{a} + \frac{x}{a - x}}{\frac{a + x}{a} - \frac{x}{a + x}}$;
в) $\frac{1}{1 + \frac{1}{1 + \frac{1}{x}}}$;
г) $\frac{1}{1 - \frac{1}{1 + \frac{1}{x}}}$.
$\frac{x - \frac{yz}{y - z}}{y - \frac{xz}{x - z}} = \frac{\frac{x(y - z) - yz}{y - z}}{\frac{y(x - z) - xz}{x - z}} = \frac{(xy - xz - yz)(x - z)}{(xy - yz - xz)(y - z)} = \frac{x - z}{y - z}$
$\frac{\frac{a - x}{a} + \frac{x}{a - x}}{\frac{a + x}{a} - \frac{x}{a + x}} = \frac{\frac{(a - x)^2 + xa}{a(a - x)}}{\frac{(a + x)^2 - ax}{a(a + x)}} = \frac{a(a^2 - 2ax + x^2 + ax)(a + x)}{a(a^2 + 2ax + x^2 - ax)(a - x)} = \frac{(a + x)(a^2 - ax + x^2)}{(a - x)(a^2 + ax + x^2)} = \frac{a^3 + x^3}{a^3 - x^3}$
$\frac{1}{1 + \frac{1}{1 + \frac{1}{x}}} = \frac{1}{1 + \frac{1}{\frac{x + 1}{x}}} = \frac{1}{1 + \frac{x}{x + 1}} = \frac{1}{\frac{x + 1 + x}{x + 1}} = \frac{x + 1}{2x + 1}$
**г
$\frac{1}{1 - \frac{1}{1 + \frac{1}{x}}} = \frac{1}{1 - \frac{1}{\frac{x + 1}{x}}} = \frac{1}{1 - \frac{x}{x + 1}} = \frac{1}{\frac{x + 1 - x}{x + 1}} = \frac{x + 1}{1} = x + 1$
Пожауйста, оцените решение