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denissavchenko
@denissavchenko
July 2022
2
22
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[tex] \sqrt{ 2^{x} \sqrt[3]{4^x \sqrt[x]{0.125} } } = 4 \sqrt[3]{2} [/tex]
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Universalka
Verified answer
1) корень x-вой степени из 0,125 = 1/8^1/x = (2^(-3))^1/x = 2^(-3/x)
2) 4^x * 2^(-3/x) = 2^(2x) * 2^(-3/x) = 2^(2x - 3/x) = 2^((2x² - 3)/x)
3) ³√2^((2x² - 3)/x) = [2^((2x² -3)/x)]^1/3 = 2^((2x² -3)/3x)
4) 2^x * 2^((2x²-3)/3x) = 2^((5x² - 3)/3x)
5) √2^((5x²-3)/3x) = [2^((5x²-3)/3x)]^1/2 = 2^((5x²-3)/6x)
6) 2^((5x²-3)/6x) = 4 ³√2
2^((5x²-3)/6x) = 2² * 2^1/3
2^((5x²-3)/6x) = 2^7/3
(5x² - 3)/6x = 7/3
5x² - 3 = 14x
5x² - 14x - 3 = 0
D/4 = 7² + 15 = 49 + 15 = 64
X1,2 = (7 + - 8)/5
X1 = 3
X2 = - 0,2
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denissavchenko
Спасибо)
ВладимирБ
Verified answer
Решение смотри на фото
2 votes
Thanks 0
denissavchenko
Cпасибо огромное)
denissavchenko
Подскажите, а почему в 4-й строке степень 7/3, а не 2/3 ???
denissavchenko
Извините, уже разобрались)
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Answers & Comments
Verified answer
1) корень x-вой степени из 0,125 = 1/8^1/x = (2^(-3))^1/x = 2^(-3/x)2) 4^x * 2^(-3/x) = 2^(2x) * 2^(-3/x) = 2^(2x - 3/x) = 2^((2x² - 3)/x)
3) ³√2^((2x² - 3)/x) = [2^((2x² -3)/x)]^1/3 = 2^((2x² -3)/3x)
4) 2^x * 2^((2x²-3)/3x) = 2^((5x² - 3)/3x)
5) √2^((5x²-3)/3x) = [2^((5x²-3)/3x)]^1/2 = 2^((5x²-3)/6x)
6) 2^((5x²-3)/6x) = 4 ³√2
2^((5x²-3)/6x) = 2² * 2^1/3
2^((5x²-3)/6x) = 2^7/3
(5x² - 3)/6x = 7/3
5x² - 3 = 14x
5x² - 14x - 3 = 0
D/4 = 7² + 15 = 49 + 15 = 64
X1,2 = (7 + - 8)/5
X1 = 3
X2 = - 0,2
Verified answer
Решение смотри на фото