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AnastasiaMi311
@AnastasiaMi311
March 2022
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Единорожек34
1). 2^(2x - 1) - 7*2^(x - 1) + 5 ≤ 0
1/2*2^(2x) - 7/2 *2^x + 5 ≤ 0
2^x = t
1/2t² - 7/2t + 5 ≤ 0 |*2
t² - 7t + 10 ≤ 0
t² - 2t - 5t + 10 ≤ 0
t(t - 2) - 5(t - 2) ≤ 0
(t - 2)(t - 5) ≤ 0
t = 2^x
(2^x - 2)(2^x - 5) ≤ 0
2^x - 2 = 0 ⇔ x = 1
2^x - 5 = 0 ⇔ x = log₂5
_+__1__-___log₂5__+__>
x∈[1 ; log₂5]
2). log₂²x + 5log₂x + 6 > 0
ОДЗ: x > 0
log₂x = t
t² + 5t + 6 > 0
t² + 2t + 3t + 6 > 0
t(t + 2) + 3(t + 2) > 0
(t + 2)(t + 3) >0
t = log₂x
(log₂x + 2)(log₂x + 3) > 0
log₂x + 2 = 0 ⇔ x = 1/4
log₂x + 3 = 0 ⇔ x = 1/8
__+__(1/8)__-___(1/4)__+___
x∈(0; 1/8)∪(1/4; ∞)
1 votes
Thanks 2
AnastasiaMi311
А что это за пробелы в конце?
Единорожек34
Координатная прямая... простите, оформить по другому не было возможности
AnastasiaMi311
А все поняла,огромное спасибо!)
Единорожек34
В первом неравенстве точки закрашены, а во втором незакрашены, иметь ввиду
AnastasiaMi311
Хорошо!
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Answers & Comments
1/2*2^(2x) - 7/2 *2^x + 5 ≤ 0
2^x = t
1/2t² - 7/2t + 5 ≤ 0 |*2
t² - 7t + 10 ≤ 0
t² - 2t - 5t + 10 ≤ 0
t(t - 2) - 5(t - 2) ≤ 0
(t - 2)(t - 5) ≤ 0
t = 2^x
(2^x - 2)(2^x - 5) ≤ 0
2^x - 2 = 0 ⇔ x = 1
2^x - 5 = 0 ⇔ x = log₂5
_+__1__-___log₂5__+__>
x∈[1 ; log₂5]
2). log₂²x + 5log₂x + 6 > 0
ОДЗ: x > 0
log₂x = t
t² + 5t + 6 > 0
t² + 2t + 3t + 6 > 0
t(t + 2) + 3(t + 2) > 0
(t + 2)(t + 3) >0
t = log₂x
(log₂x + 2)(log₂x + 3) > 0
log₂x + 2 = 0 ⇔ x = 1/4
log₂x + 3 = 0 ⇔ x = 1/8
__+__(1/8)__-___(1/4)__+___
x∈(0; 1/8)∪(1/4; ∞)