[latex] \frac{3- 4^{x} }{2- 2^{x} } \geq 1,5 [/latex]

Пусть [latex]y=2^x[/latex]. Тогда неравенство верно при [latex]\frac{3-y^2}{2-y} \geq \frac{3}{2} \Rightarrow \frac{2(3-y^2)-3(2-y)}{2(2-y)} \geq 0[/latex] Тогда [latex]\frac{-2y^2+3y}{4-2y} \geq 0 \Rightarrow \left \{ {{y(3-2y) \geq 0} \atop {4-2y \ \textgreater \ 0}} \right. or \left \{ {{y(3-2y) \leq 0} \atop {4-2y \ \textless \ 0}} \right.[/latex]  [latex] \left \{ {{y(3-2y) \geq 0} \atop {4-2y \ \textgreater \ 0}} \right. \Rightarrow y \in (-\infty,2) \cap ([0,+\infty)\cap(-\infty,\frac{3}{2}] \cup(-\infty,0]\cap[\frac{3}{2},\infty))[/latex] [latex]\Rightarrow y \in [0,\frac{3}{2}][/latex] [latex]\left \{ {{y(3-2y) \leq 0} \atop {4-2y \ \textless \ 0}} \right. \Rightarrow y \in (2,+\infty) \cap ((-\infty,0] \cup [\frac{3}{2},+\infty)) \Rightarrow y \in (2,\infty)[/latex] Тогда [latex]y \in [0,\frac{3}{2}] \cup (2,\infty) \Rightarrow x \in (-\infty, \frac{ln(1.5)}{ln(2)}] \cup (1, \infty)[/latex]