Question

Пожалуйста, решите неравенства!!!!
x+1<$frac{x}{2}$
2≤x²+x<6

x⁶-9x³+8<0

Answer 1

$x+1 textless frac x2\ 2x+2 textless x\ 2x-x textless -2\ x textless -2\\ 2leq x^2+x textless 6\ begin{cases}x^2+xgeq2\x^2+x textless 6end{cases}Rightarrowbegin{cases}x^2+x-2geq0\x^2+x-6 textless 0end{cases}Rightarrowbegin{cases}(x+5)(x-4)geq0\(x+3)(x-2) textless 0end{cases}\xin(-3;;2)\\x^6-9x^3+8 textless 0\x^3=t,;x^6=t^2,;t textgreater 0\t^2-9t+8 textless 0\(t-1)(t-8) textless 0\(x^3-1)(x^3-8) textless 0\(x^3-1)(x^3-8)=0\x_1=1,;x_2=2\xin(1;;2)$