• Предмет: Алгебра
  • Автор: olyaberdnikova1
  • Вопрос задан 8 лет назад

помогите срочно. Зарание спасибо

Приложения:

Ответы

Ответ дал: maymr
0
В1.
 frac{a {}^{3} }{a {}^{2} - 4 }  +  frac{a}{2 - a}  -  frac{2}{a + 2} - a + 4 =  \  \  =  frac{ {a}^{3} }{(a - 2)(a + 2)}   -  frac{a}{a - 2}  -  frac{2}{a + 2}  -  frac{a {}^{3}  - 4a}{ {a}^{2} - 4 }  +  frac{4 {a}^{2}  - 16}{ {a}^{2} - 4 }  =  \  \  =  frac{ {a}^{3} }{ {a}^{2}  - 4}  -  frac{a {}^{2} + 2a }{a {}^{2}  - 4}  -  frac{2a - 4}{ {a}^{2}  - 4}  -  frac{a {}^{3}  - 4a}{ {a}^{2}  - 4}  +  frac{4 {a}^{2} - 16 }{ {a}^{2}  - 4}  =  \  \  =  frac{ {a}^{3} -  {a}^{2} - 2a - 2a + 4 -  {a}^{3}   + 4a + 4 {a}^{2} - 16  }{ {a}^{2}  - 4}  =  \  \  =  frac{3 {a}^{2} - 12 }{ {a}^{2} - 4 }  =  \  \  =  frac{3(a {}^{2}  - 4)}{ {a}^{2}  - 4}  =  \  \  =  frac{3(a - 2)(a + 2)}{(a - 2)(a + 2)}  =  \  \  =  3
Ответ: 3

С1.
 frac{1}{x(x + 1)}  +  frac{1}{(x + 1)(x + 2)}  +  frac{1}{(x + 2)(x + 3)}  -  frac{3}{x(x + 3)}  - 3 =  \  \  =  frac{x {}^{2}  + 5x + 6 + x {}^{2}  + 3x + x {}^{2}  + x - 3x {}^{2}  -  9x  -  6 - 3(x {}^{2}   + x)(x {}^{2}  + 5x + 6)}{x(x + 1)(x + 2)(x + 3)}  =  \  \  =  frac{ - 3(x {}^{4}  + 5x {}^{3}  + 6x {}^{2}  + x {}^{3}  + 5x {}^{2}  + 6x)}{x(x + 1)(x + 2)(x + 3)}  =  \  \  =  frac{ - 3x {}^{4} - 15x {}^{3} - 18x {}^{2}  - 3x {}^{3}  - 15x {}^{2}  - 18x  }{x {}^{4}  + 6x {}^{3}  + 11x {}^{2}   + 6x  }  =  \  \  =  frac{ - 3x {}^{4}  - 18x {}^{3} - 33x {}^{2}   - 18x}{x {}^{4} + 6x {}^{3} + 11x {}^{2}   + 6x }  =  \ \   =   frac{ - x(3x {}^{3} + 18x {}^{2} + 33x + 18)  }{x(x {}^{3} + 6x {}^{2} + 11x + 6)  }  =  \  \  =  frac{ - (3x {}^{3}  + 33x)(18x {}^{2} + 18) }{(x {}^{3} + 11x)(6x {}^{2}  + 6) }  =  \  \  =  -  frac{3x(x {}^{2}  + 11) + 18(x {}^{2} + 1) }{x(x {}^{2}  + 11) + 6(x {}^{2} + 1) }  =  \  \  =  -  frac{3x + 18}{x + 6}  =  \  \  =  -  frac{3(x + 6)}{x + 6}  =  \  \  =  - 3
Ответ: -3
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