Question

folosin monotonia functiei exponentiale sa se rezolve inecuatia
(√2-1)^(x-x^2)>=(√2+1)^x

Answer 1

$\displaystyle \\ \\ Observam~ca~( \sqrt{2}-1)( \sqrt{2}+1)=2-1=1 \Rightarrow \sqrt{2}-1= \frac{1}{ \sqrt{2}+1}= \\ \\ =(\sqrt{2}+1)^{-1} \\ \\ ( \sqrt{2}-1)^{x-x^2} \ge (\sqrt{2}+1)^x \Leftrightarrow \\ \\ \Leftrightarrow \left( (\sqrt{2}+1)^{-1} \right)^{x-x^2} \ge ( \sqrt{2}+1)^x \Leftrightarrow \\ \\ \Leftrightarrow (\sqrt2+1)^{x^2-x} \ge ( \sqrt{2}+1)^x. \\ \\ Deoarece~ \sqrt{2}+1\ \textgreater \ 1,~rezulta~x^2-x \ge x \Leftrightarrow x(x-2) \ge 0. \\ \\ Solutie:~x \in (- \infty ; 0] \cup [2;+ \infty).$