показателно неравенство

[Гост]

[tex]\frac{9}{2^{x}- 1} - 2 \ge \frac{21}{2^{x}+ 1}[/tex]

[ammornil]

[tex]\frac{9}{2^{x}- 1} - 2 \ge \frac{21}{2^{x}+ 1}\\[/tex]

[tex]\begin{array}{llll} &&& \text{ДМ: } 2^{x}-1\ne 0 \Rightarrow x \ne 0 \\ 2^{x}=u > 0 \\ \frac{\normalsize{9}}{\normalsize{u- 1}} - 2- \frac{\normalsize{21}}{\normalsize{u+ 1}} \ge 0 \\ 9(u+1)-2(u+1)(u-1) -21(u-1) \ge 0 \\ 9u+9-2u^{2}+2-21u+21 \ge 0 \\ 2u^{2}+12u-32 \le 0 & D=6^{2}-2\cdot{(-32)}=100 & u_{1,2}=\frac{-6\pm10}{2} \\ 2(u+8)(u-2)\le 0 \\ 2(2^{x}+8)(2^{x}-2)\le 0 \cap 2^{x} > 0 \\ 2^{x} \in (0;2] \\ x \ne 0 \Rightarrow x \in (-\infty; 0) \cup (0;1]\end{array}[/tex]