показателно неравенство

[Гост]

[tex]2^{2x+1}-5\cdot6^{x}+3^{2x+1} \ge 0[/tex]

[ammornil]

[tex]2^{2x+1}-5\cdot6^{x}+3^{2x+1} \ge 0[/tex]

[tex]\begin{array}{llll}&&\text{ДМ: } \forall{x}\in\mathbb{R} \\ & 2^{2x}\cdot{2^{1}}-5\cdot{(2\cdot{3})^{x}}+3^{2x}\cdot{3^{1}}\ge{0} \\ & 2\cdot{2^{2x}}-5\cdot{2^{x}}\cdot{3^{x}}+3\cdot{3^{2x}}\ge{0} \hspace{0.8em}| \div{} 2^{2x}\ne{0} \\ & 2\cdot{1}-5\cdot{\frac{\normalsize{3^{x}}}{\normalsize{2^{x}}}}+3\cdot{\frac{\normalsize{3^{2x}}}{\normalsize{2^{2x}}}}\ge 0 \\ & 3\cdot{\left(\frac{\normalsize{3^{x}}}{\normalsize{2^{x}}}\right)^{2}} -5\cdot{\frac{\normalsize{3^{x}}}{\normalsize{2^{x}}}}+2\ge{0} \\ u=\frac{\normalsize{3^{x}}}{\normalsize{2^{x}}}>0 \Rightarrow & 3u^{2}-5u+2\ge{0} & D=(-5)^{2}-4\cdot{3}\cdot{2}=1 \\ && u_{1,2}=\frac{-(-5)\pm1}{2\cdot{3}} \Leftrightarrow \begin{cases} u_{1}=1 \in \text{Дu} \\ u_{2}=\frac{\normalsize{2}}{\normalsize{3}} \in \text{Дu} \end{cases} \\ & & \begin{array}{lcl} \frac{\normalsize{3^{x}}}{\normalsize{2^{x}}}=1 &\cup & \frac{\normalsize{3^{x}}}{\normalsize{2^{x}}}=\frac{\normalsize{2}}{\normalsize{3}} \\ \left(\frac{\normalsize{3}}{\normalsize{2}}\right)^{x} = \left(\frac{\normalsize{3}}{\normalsize{2}}\right)^{0} & \cup & \left(\frac{\normalsize{3}}{\normalsize{2}}\right)^{x}=\left(\frac{\normalsize{3}}{\normalsize{2}}\right)^{-1} \\ x_{1}=0 & \cup & x_{2}=-1 \end{array} \\ & 3\left(u-\frac{\normalsize{2}}{\normalsize{3}} \right)(u-1)\ge 0 \\ & u \in \left(-\infty ; \frac{\normalsize{2}}{\normalsize{3}} \right] \cup [1; +\infty) \\ & x \in (-\infty;-1] \cup [0, +\infty) \end{array}[/tex]