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)
$\frac{x^3-x^2+x-1}{x+8}\leq0, \\ \left \{ {{x+8\neq0,} \atop {(x^3-x^2+x-1)(x+8)\leq0;}} \right. \\ x\neq-8; \\ (x^2(x-1)+x-1)(x+8)\leq0, \\ (x^2(x-1)+x-1)(x+8)\leq0, \\ (x-1)(x^2+1)(x+8)\leq0, \\ x^2+1\ \textgreater \ 0 \ \ \forall x\in R, \\ (x-1)(x+8)\leq0, \\$
$\left [ {{ \left \{ {{x-1\leq0,} \atop {x+8\ \textgreater \ 0;}} \right. } \atop { \left \{ {{x-1\geq0,} \atop {x+8\ \textless \ 0;}} \right. }} \right. \left [ {{ \left \{ {{x\leq1,} \atop {x\ \textgreater \ -8;}} \right. } \atop { \left \{ {{x\geq1,} \atop {x\ \textless \ -8;}} \right. }} \right. \left [ {{-8\ \textless \ x\leq1,} \atop {x\in\varnothing;}} \right. \\ x\in(-8;1].$
Б)
$\frac{x^3-11x^2+39x-45}{x+2)}\geq0, \\ \left \{ {{x+2\neq0,} \atop {(x^3-11x^2+39x-45)(x+2)\geq0;}} \right. \\ x\neq-2, \\ (x^3-3x^2-8x^2+24x+15x-45)(x+2)\geq0, \\ (x^2(x-3)-8x(x-3)+15(x-3))(x+2)\geq0, \\(x^2-8x+15)(x-3)(x+2)\geq0, \\ x^2-8x+15=0, x_1=3, x_2=5, \\ (x-3)(x-5)(x-3)(x+2)\geq0, \\$
$(x-3)^2(x-5)(x+2)\geq0, \\ (x-3)^2\geq0 \ \ \forall x\in R, \\ (x-5)(x+2)\geq0, \\ \left [ {{ \left \{ {{x-5 \geq 0,} \atop {x+2\ \textgreater \ 0,}} \right. } \atop { \left \{ {{x-5 \leq 0,} \atop {x+2\ \textless \ 0;}} \right. }} \right. \left [ {{ \left \{ {{x\geq 5,} \atop {x\ \textgreater \ -2,}} \right. } \atop { \left \{ {{x\leq 5,} \atop {x\ \textless \ -2;}} \right. }} \right. \left [ {{x \geq 5,} \atop {x\ \textless \ -2;}} \right. \\x\in(-\infty;-2)\cup[5;+\infty)\cup\{3\}.$