Question

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

Answer 1

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

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

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

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

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