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erenyeger2398
erenyeger2398
12.04.2020 04:14 •  Математика

2в степени 3x равное 512 в степени 1 разделить на 3x

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Ответ:
TheMissteress
TheMissteress
12.04.2020
2^{3x}=512^{ \frac{1}{3x}}\\
2^{3x}=2^9^{ \frac{1}{3x}}\\}\\
3x= \frac{9}{3x}\\
x=1 ; x=-1
4,6(2 оценок)
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Ответ:
ROMMIT2
ROMMIT2
12.04.2020

Пошаговое объяснение:

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4,4(68 оценок)
Ответ:
katyaarxipova
katyaarxipova
12.04.2020
В заданном уравнении 2cos^2 x - 3√3 cos x + 3 < 0 произведём замену: cos x = у.
Получаем неравенство:
2у² - 3√3у + 3 <0.
Графически - это часть параболы у = 2у² - 3√3у + 3, расположенная ниже оси х.
Находим граничные точки - точки пересечения параболы с оью х. То есть решаем квадратное уравнение:
2у² - 3√3у + 3 = 0.
Квадратное уравнение, решаем относительно y: 
Ищем дискриминант:D=(-3*2root3)^2-4*2*3=9*3-4*2*3=27-4*2*3=27-8*3=27-24=3;
Дискриминант больше 0, уравнение имеет 2 корня:y_1=(2root3-(-3*2root3))/(2*2)=(√3+3*2√3)/(2*2)=4*√3/(2*2)=4*√3/4=√3≈1.7321; (этот корень отбрасываем - больше 1);y_2=(-√3-(-3*√3))/(2*2)=(-√3+3*√3)/(2*2)=2*√3/(2*2)=2*√3/4=√3/2 ≈0.8660.
Обратная замена: cos x = √3/2.
Получили предельные значения х.
х = (-π/6)+2πk, k ∈ Z.
x = (π/6)+2πk, k ∈ Z. 

Тогда ответ: (-π/6)+2πk < x < (π/6)+2πk
4,7(25 оценок)
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