Question

Решите равенство:
х^3-х^2+х-1
А) =<0
Х+8

Х^3-11х^2+39х-45
Б) >=0
Х+2

Answer 1

А)
$frac{x^3-x^2+x-1}{x+8}leq0, \ left { {{x+8neq0,} atop {(x^3-x^2+x-1)(x+8)leq0;}} right. \ xneq-8; \ (x^2(x-1)+x-1)(x+8)leq0, \ (x^2(x-1)+x-1)(x+8)leq0, \ (x-1)(x^2+1)(x+8)leq0, \ x^2+1 textgreater 0 forall xin R, \ (x-1)(x+8)leq0,$
$left [ {{ left { {{x-1leq0,} atop {x+8 textgreater 0;}} right. } atop { left { {{x-1geq0,} atop {x+8 textless 0;}} right. }} right. left [ {{ left { {{xleq1,} atop {x textgreater -8;}} right. } atop { left { {{xgeq1,} atop {x textless -8;}} right. }} right. left [ {{-8 textless xleq1,} atop {xinvarnothing;}} right. \ xin(-8;1].$
Б)
$frac{x^3-11x^2+39x-45}{x+2)}geq0, \ left { {{x+2neq0,} atop {(x^3-11x^2+39x-45)(x+2)geq0;}} right. \ xneq-2, \ (x^3-3x^2-8x^2+24x+15x-45)(x+2)geq0, \ (x^2(x-3)-8x(x-3)+15(x-3))(x+2)geq0, \(x^2-8x+15)(x-3)(x+2)geq0, \ x^2-8x+15=0, x_1=3, x_2=5, \ (x-3)(x-5)(x-3)(x+2)geq0,$
$(x-3)^2(x-5)(x+2)geq0, \ (x-3)^2geq0 forall xin R, \ (x-5)(x+2)geq0, \ left [ {{ left { {{x-5 geq 0,} atop {x+2 textgreater 0,}} right. } atop { left { {{x-5 leq 0,} atop {x+2 textless 0;}} right. }} right. left [ {{ left { {{xgeq 5,} atop {x textgreater -2,}} right. } atop { left { {{xleq 5,} atop {x textless -2;}} right. }} right. left [ {{x geq 5,} atop {x textless -2;}} right. \xin(-infty;-2)cup[5;+infty)cup{3}.$

Answer 2

Огромное спасибо вам!

Answer 3

ничего не пончтно