Question Number 198228 by sulaymonnorboyev140 last updated on 14/Oct/23
$$\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)^{{x}^{\mathrm{2}} −{x}} >\mathrm{1} \\ $$
Answered by MM42 last updated on 14/Oct/23
$$\begin{cases}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}>\mathrm{1}}\\{{x}^{\mathrm{2}} −{x}>\mathrm{0}}\end{cases}\Rightarrow\left(−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\mathrm{0},+\infty\right)={A}\:\:\&\:\left(−\infty,\mathrm{0}\right)\cup\left(\mathrm{1},+\infty\right)={B} \\ $$$$\Rightarrow{A}\cap{B}=\left(−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\mathrm{1},+\infty\right)\:\left({i}\right) \\ $$$$\begin{cases}{\mathrm{0}<\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}<\mathrm{1}}\\{{x}^{\mathrm{2}} −{x}<\mathrm{0}}\end{cases}\Rightarrow\left(−\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0}\right)={C}\:\:\&\:\left(\mathrm{0},\mathrm{1}\right)={D} \\ $$$$\Rightarrow{C}\cap{D}=\phi\:\left({ii}\right) \\ $$$$\left({i}\right)\cup\left({ii}\right)=\left({i}\right) \\ $$$$ \\ $$$$ \\ $$