Question

помогите решить 5 задание​

Answer 1

Объяснение:

ОДЗ:

$x\in\mathbb N,\ \ \ \ x\geq 3.$

$C_x^1+6*C_x^2+6*C_x^3\leq 9x^2-14x\\\frac{x!}{(x-1)!*1!}+\frac{6*x!}{(x-2)!*2!} +\frac{6*x!}{(x-3)!*3!} \leq 9x^2-14x\\\frac{(x-1)!*x}{(x-1)!*1}+\frac{6*(x-2)!*(x-1)*x}{(x-2)!*1*2} +\frac{6*(x-3)!*(x-2)*(x-1)*x}{(x-3)!*1*2*3} \leq 9x^2-14x\\x+3*(x-1)*x+(x-2)*(x-1)*x\leq 9x^2-14x\\x+x*(3x-3)+x*(x^2-3x+2)-9x^2+14x\leq 0\\x*(1+3x-3+x^2-3x+2-9x+14)\leq 0\\x*(x^2-9x+14)\leq 0\\x*(x^2-7x-2x+14)\leq 0\\x*(x*(x-7)-2*(x-7))\leq 0\\x*(x-7)*(x-2)\leq 0$

-∞__-__0__+__2__-__7__+__+∞        ⇒

x∈(-∞;0]U[2;7].

Согласно ОДЗ:

Ответ: x=3, 4, 5. 6, 7.