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Опубликовано 5 лет назад по предмету Алгебра от Poli44444

Решите систему неравенств под буквой Б

  1. Ответ
    Ответ дан arsenlevadniy
     left { {{x^2-x-12 textless  0,} atop {x^2-4x+3 geq 0;}} right.  left { {{(x+3)(x-4) textless  0,} atop {(x-1)(x-3)geq 0;}} right. left { {{-3 textless  x textless  4,} atop {left { {{x leq 1,} atop {x geq 3;}} right.}} right.  left [ {{-3 textless  x leq 1,} atop {3 leq x textless  4;}} right. \ xin(-3;1]cup[3;4).

     left { {{x^2+x-2 geq 0,} atop {frac{x+2}{x-4}leq 0}} right.   left { begin{array}{ccc}(x+2)(x-1) geq 0,\x-4 neq 0,\(x+2)(x-4) leq 0;end{array}right.  left { begin{array}{ccc} left [ {{x leq -2,} atop {x geq 1,}} right. \x neq 4,\-2 leq x leq 4;end{array}right.  left [ {{x=-2,} atop {1 leq x textless  4;}} right. \ xin{-2}cup[1;4).

     left { {{|x| leq 2,} atop {x^2-x-6 geq 0;}} right.  left { {{-2 leq x leq 2,} atop {(x+2)(x-3) geq 0;}} right.  left { {{-2 leq x leq 2,} atop {(x+2)(x-3) geq 0;}} right. left { {{-2 leq x leq 2,} atop { left [ {{x leq -2,} atop {x geq 3;}} right. }} right. \ x=-2.
    1. Ответ
      Ответ дан Poli44444
      спасибо Вам большое!