Решите неравенство 2^−2√x + 32 ⋅ 10^2−√x > 2^9−2√x + 625 ⋅ 10^−2−√x

Источник: ЕГЭ ОГЭ по математике Под редакцией И.В. Ященко.

Решение:

2^−2√x + 32 ⋅ 10^2−√x > 2^9−2√x + 625 ⋅ 10^−2−√x

2^−2√x + 32 ⋅ 10^2+(-√x) > 2^9+(-2√x) + 625 ⋅ 10^−2+(-√x)

2^−2√x + 32 ⋅ 10^-√x > 2^-2√x + 625 ⋅ 1/10^-√x

2^−2√x + 3200 ⋅ 5^-√x > 512 ⋅ 2^-√x + 25/4 ⋅ 5^-√x

3200 ⋅ 5^-√x – 25/4 ⋅ 5 ^-√x > 512 ⋅ 2 ^-√x + 25/4 ⋅ 5^-√x

5² 5 ^-√x ⋅ (128 – 1/4) > 511 ⋅ 2^-√x

5² 5 ^-√x ⋅ (512 – 1) > 511 ⋅ 2^-√x ⋅ 2²

5² 5 ^-√x > 2 ^-√x ⋅ 2²

(5/2) ^-√x > (5/2)^-2

^-√x > -2

^√x < 2

0 < x < 4

Ответ: 0;4