Предмет: Алгебра, автор: lukasoe

                                 

решить уравнение 2 в степени х + 2 в степени 2-х     =5

 

найти корень уравнения cos2x+3sinx=2

                                                                                            

решить уравнение  корень(2-x)/(x+3) + корень(x+3)/(2-x) =3 1/3

                                                                                            

Ответы

Автор ответа: vajny
0

1) Обозначим 2 в степ х = у.

Тогда у+(4/у) = 5, или

у квад - 5у + 4 = 0. Корни по т, Виета:

у1=1; у2 = 4. Значит имеем два уравнения:

2 в степ х = 1, здесь х = 0, и

2 в степ х = 4, здесь х = 2.

Ответ: 0; 2.

 

2) Преобразуем с использованием формулы cos двойного угла:

1 - 2sin квад х + 3 sinx = 2,

2у квад - 3у + 1 = 0, где у = sinx принадл. [-1;1]

D = 1, у1 = 1/2; у2 = 1,

Или sin x = 1/2, x= (-1)в степ k *П/6   +   Пk, или

       sin x = 1, x = П/2 + 2Пk.

Ответ: (-1)в ст.k * П/6   +   Пk;   П/2 +2Пk.  k прин. Z.

 

3) Найдем ОДЗ: (2-х)/(х+3) больше 0. Методом интервалов получим допустимую область для х:

(-3; 2).

Обозначим первый из корней за у, причем у больше 0. Тогда:

у + (1/у) = 10/3, или:

3у квад - 10у + 3 = 0, D = 64. Тогда:

у1 = 1/3, у2 = 3.

Решаем:

кор[(2-x)/(x+3)] = 1/3.   18 - 9x = x + 3, x = 1,5  - входит в ОДЗ. Теперь решаем:

кор[(2-x)/(x+3)] = 3.    9х + 27 = 2 - х, х = - 2,5  - входит в ОДЗ.

 

Ответ: - 2,5; 1,5.

 

 

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