Question

Решить неравенство $x^{3} - x + sqrt[]{ x^{3} -x+1} >= 1$ .

Answer 1

$t=x^3-x\\t+sqrt{t+1} geq 1\\sqrt{t+1} geq 1-t\\t+1 geq 1-2t+t^2\\t^2-3t leq 0\\t(t-3) leq 0; ; ; ++++[0]----[3]++++\\0 leq t leq 3\\ left { {{x^3-x geq 0} atop {x^3-x leq 3}} right. left { {{x(x-1)(x+1) geq 0} atop {x^3-x-3 leq 0}} right. ; left { {{xin [-1,0]U[1,infty)} atop {x^3-x-3leq 0}} right.$

$x^3-x-3=0\\x_1=sqrt[3]{frac{3}{2}-sqrt{frac{9}{4}-frac{1}{27}}}+sqrt[3]{frac{3}{2}+sqrt{frac{9}{4}-frac{1}{27}}}approx 1,67\\x^3-x-3 leq 0,x leq x_1,x leq 1,67\\Otvet:xin [-1,0]U[1;1,67]$