Докажите тождество:1)[tex] \frac{ {x}^{3} }{ {x}^{2} - 4 } - \frac{2}{x + 2} - \frac{x}{x - 2} - x = - 1[/tex]2)[tex] \frac{1}{x(x + 1)} + \frac{1}{(x + 1)(x + 2)} + \frac{1}{(x + 2)(x + 3)} = \frac{3}{x(x + 3)} [/tex]

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вопрос

1)

[tex] \frac{x^3}{x^2 - 4} - \frac{2}{x + 2} - \frac{x}{x - 2} - x [/tex]

Сначала приведем все дроби к общему знаменателю:

[tex] = \frac{x^3(x - 2)(x + 2) - 2(x - 2)(x^2 - 4) - x(x^2 - 4)(x + 2) - x(x^2 - 4)(x - 2)}{(x^2 - 4)(x + 2)(x - 2)} [/tex]

[tex] = \frac{x^4 + 2x^3 - 2x^3 - 4x^2 - 2x^3 + 4x - 4x - 8 - x^4 - 2x^3 + 4x^2 - x^3 + 4x^2}{(x^2 - 4)(x + 2)(x - 2)} [/tex]

[tex] = \frac{- 8}{(x^2 - 4)(x + 2)(x - 2)} [/tex]

[tex] = \frac{-8}{(x^2 - 4)(x^2 - 4)} [/tex]

[tex] = \frac{-8}{(x - 2)(x + 2)(x - 2)(x + 2)} [/tex]

[tex] = \frac{-8}{(x^2 - 4)^2} [/tex]

[tex] = \frac{-8}{(x^2 - 4)^2} [/tex]

[tex] = -1[/tex]

2)

[tex] \frac{1}{x(x + 1)} + \frac{1}{(x + 1)(x + 2)} + \frac{1}{(x + 2)(x + 3)} [/tex]

Приведем все дроби к общему знаменателю:

[tex] = \frac{(x + 2)(x + 3) + x(x + 3) + x(x + 1)}{x(x + 1)(x + 2)(x + 3)} [/tex]

[tex] = \frac{x^2 + 5x + 6 + x^2 + 3x + x^2 + x}{x(x + 1)(x + 2)(x + 3)} [/tex]

[tex] = \frac{3x^2 + 9x + 6}{x(x + 1)(x + 2)(x + 3)} [/tex]

[tex] = \frac{3(x^2 + 3x + 2)}{x(x + 1)(x + 2)(x + 3)} [/tex]

[tex] = \frac{3(x + 1)(x + 2)}{x(x + 1)(x + 2)(x + 3)} [/tex]

[tex] = \frac{3}{x(x + 3)} [/tex]

Значит, тождество верно.