Question

Решите неравенство: 2^x+2 − 24 / 2^x+1 − 8 ≥ 1

Answer 1

Объяснение:

ОДЗ: 2ˣ⁺¹-8≠0     2*2ˣ≠8 |÷2     2ˣ=4     2ˣ≠2²     x≠2.

$\frac{2^{x+2}-24}{2^{x+1}-8}\geq 1\ \ \ \frac{2^2*2^x-24}{2*2^x-8 } \geq 1\ \ \ \frac{4*2^x-24}{2*(2^x-4) } \geq 1\ \ \ \frac{4*(2^x-6)}{2*(2^x-4)}\geq 1\ \ \ \frac{2*(2^x-6)}{2^x-4} -1\geq 0\\\frac{2*2^x-12-2^x+4}{2^x-4} \geq 0\ \ \ \frac{2^x-8}{2^x-4}\geq 0\ \ \ \frac{2^x-2^3}{2^x-2^2} \geq 0$

$\left \{ {{2^x-2^3=0} \atop {2^x-2^2=0}} \right.\ \ \ \left \{ {{2^x=2^3} \atop {2^x=2^2}} \right. \ \ \ \left \{ {{x=3} \atop {x=2}} \right. \Rightarrow$

-∞__+__2__-__3__+__+∞

Согласно ОДЗ:

ответ: x∈(-∞;2)U[3;+∞).