Задание 1263

$$\frac{7,5-4\cdot 2^x }{4^x-10\cdot 2^x+16 } \geqslant 0,5$$

ОДЗ: $4^x-10\cdot 2^x+16 \neq 0\Rightarrow$$x\neq 3, x\neq 1$

$$2^{2x}-10\cdot 2^x+16 \neq 0$$

Пусть $2^x=t$

$$t^2-10t+16\neq 0$$

$D=(-10)^2-4\cdot 16=$$100-64=36$

$t_1=\frac{-(-10)+\sqrt{36} }{2}=\dfrac{10+6}2=$$\dfrac{16}2=8$

$$t_2=\dfrac{10-6}2=\dfrac42=2$$

$$2^x \neq 8$$

$$2^x \neq 2^3$$

$$x\neq 3$$

$$2^x \neq 2$$

$$2^x \neq 2^1$$

$$x\neq 1$$

$$\frac{7,5-4t }{(t-8)(t-2)}-\dfrac12 \geqslant 0$$

$\frac{2\cdot (7,5-4t)-(t^{2}-10t+16)}{2(t-8)(t-2)}$$\geqslant 0$

$$\frac{15-8t-t^2+10t-16}{2(t-8)(t-2)} \geqslant 0$$

$$\frac{-t^2+2t-1}{2(t-8)(t-2) } \geqslant 0$$

$$\frac{t^2-2t+1}{2(t-8)(t-2)} \leqslant 0$$

$$\frac{(t-1)^2 }{2(t-8)(t-2)} \leqslant 0$$

$$1)t=1$$

$$2^x=1$$

$$2^x=2^0$$

$$x=0$$

$$2) 2< t < 8$$

$$2 < 2^x < 8$$

$$2^2 < 2^x < 2^3$$

$$2< x < 3$$