Question

решите неравенство (y-1)(y-3)(y-4)(y-6)-9>=0
Помогите пожалуйста )

Answer 1

$(y-1)(y-3)(y-4)(y-6)-9geq 0\((y-1)(y-6))*((y-3)(y-4))geq 0\(y^2-6y-y+6)(y^2-4y-3y+12)-9geq 0\(y^2-7y+6)(y^2-7y+12)-9geq 0\y^2-7y=t\(t+6)(t+12)-9geq 0\t^2+12t+6t+72-9geq 0$

$t^2+18t+63geq 0\D=18^2-4*63=9^2*4-4*9*7=9*4(9-7)=9*4*2=(6sqrt{2})^2\t_1=frac{-18+6sqrt{2}}{2} =3sqrt{2}-9\t_2=frac{-18-6sqrt{2}}{2} =-3sqrt{2}-9\+ -   +\---[]---------[]----->\ -3sqrt{2}-9   3sqrt{2}-9\t in (-infty;-3sqrt{2}-9)cup (3sqrt{2}-9;+infty)\left[begin{array}{ccc}y^2-7yleq -3sqrt{2}-9\y^2-7y geq 3sqrt{2}-9end{array}right.$

$left[begin{array}{ccc}y^2-7y+3sqrt{2}+9leq 0\y^2-7y-3sqrt{2}+9 geq 0 end{array}right.\y^2-7y+3sqrt{2}+9leq 0\D=49-12sqrt{2}-36<0 Rightarrow y^2-7y+3sqrt{2}+9>0, forall xin R\y^2-7y-3sqrt{2}+9 geq 0\D=49+12sqrt{2}-36=13+12sqrt{2}$

$y_1=frac{7+sqrt{13+12sqrt{2}}}{2} \y_2=frac{7-sqrt{13+12sqrt{2}}}{2}\+ - +\---[]--------[]--------->\frac{7-sqrt{13+12sqrt{2}}}{2} frac{7+sqrt{13+12sqrt{2}}}{2}\y in (-infty;frac{7-sqrt{13+12sqrt{2}}}{2})cup (frac{7+sqrt{13+12sqrt{2}}}{2};+infty)$

Ответ: $y in (-infty;frac{7-sqrt{13+12sqrt{2}}}{2})cup (frac{7+sqrt{13+12sqrt{2}}}{2};+infty)$