rozwiąż nierówność

[delonge]

1. \(\displaystyle{ \sqrt{2+x- x^{2}} \ge x-4}\)

[Kubica]

\(\displaystyle{ D _{f}: 2+x- x^{2} \ge 0 \\
\Delta=1+8=9 \\
x_{1} = \frac{-1+3}{-2}=-1 \\
x_{2} = \frac{-1-3}{-2}=2 \\
x \in <-1,2> \\
D _{f}=<-1,2> \\
\\
\sqrt{2+x- x^{2}} ^{2} \ge \left(x-4 \right)^{2} \\
2+x- x^{2} \ge x^{2}-8x+16 \\
2x^{2}-9x+14 \le 0 \\
\Delta=-31 \\
x \in \emptyset}\)