Question

Помогите решить уровнение!

Answer 1

Все вычисления в фото.

А)

$\frac{x}{2x {}^{2} - 18 } \times (x {}^{2} - 6x + 9) = \frac{x}{2(x { - 9)}^{2} } \times (x - 3) {}^{2} = \frac{x}{2(x {}^{2} - 9)} \times (x - 3) {}^{2} = \frac{x {}^{2} - 3x }{2x + 6} = \frac{x(x - 3)}{2(x + 3)}$

Б)

$\frac{4 + x}{4 - x} \times ( \frac{2x {}^{2} }{4 +x} - x) = \frac{4 + x}{ - (x - 4)} \times \frac{2x {}^{2} - x \times (4 + x) }{4 + x} = \frac{1}{ - (x - 4)} \times (x {}^{2} - 4x) = x$

В)

$\frac{a {}^{2} - 4 }{9 - b {}^{2} } \div \frac{a - 2}{3 - b} - \frac{2}{3 + b} = \frac{(a - 2)(a + 2)}{(3 - b)(3 + b)} \times \frac{3 - b}{a - 2} = \frac{a + 2}{3 + b}$

Г)

$( \frac{1}{y - 1} - \frac{y + 1}{y {}^{2} + y + 1 } ) \div (1 + \frac{1}{y {}^{3} - 1} ) = \frac{y {}^{2} + y + 1 - (y - 1)(y + 1)}{(y - 1)(y {}^{2} + y + 1)} \times \frac{y {}^{3} - 1}{y {}^{3} } = \frac{y + 2}{y {}^{3} }$