Question

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

x⁶-9x³+8<0

Answer 1

$x+1\ \textless \ \frac x2\\ 2x+2\ \textless \ x\\ 2x-x\ \textless \ -2\\ x\ \textless \ -2\\\\ 2\leq x^2+x \ \textless \ 6\\ \begin{cases}x^2+x\geq2\\x^2+x\ \textless \ 6\end{cases}\Rightarrow\begin{cases}x^2+x-2\geq0\\x^2+x-6\ \textless \ 0\end{cases}\Rightarrow\begin{cases}(x+5)(x-4)\geq0\\(x+3)(x-2)\ \textless \ 0\end{cases}\\x\in(-3;\;2)\\\\x^6-9x^3+8\ \textless \ 0\\x^3=t,\;x^6=t^2,\;t\ \textgreater \ 0\\t^2-9t+8\ \textless \ 0\\(t-1)(t-8)\ \textless \ 0\\(x^3-1)(x^3-8)\ \textless \ 0\\(x^3-1)(x^3-8)=0\\x_1=1,\;x_2=2\\x\in(1;\;2)$