Решите уравнения пожалуйста, ( кроме последнего) заранее спасибо. Created by meme9. algebra-ru

1. √(2x+2) =1 -x⇔ {1 -x ≥ 0 ;2x +2 = (1-x)². ⇔{ x ≤ 1 ; x² -4x -1 =0.⇔{ x ≤ 1 ; x= 2 ±√5.⇒ x= 2 -√5 .-------2.√(8+10cosx) =2sinx . ⇔{ sinx ≥0 ; 8+10cosx =4sin²x. ⇔{ sinx ≥0 ; 8+10cosx =4(1-cos²x) .⇔{ sinx ≥0 ; 2cos²x+5cosx +2=0.⇔{ sinx ≥0 ; cos²x+(2+1/2)cosx +1=0.⇔{ sinx ≥0; [ cosx= -2(н.р.);cosx = -1/2. ⇔{ sinx ≥0; cosx = -1/2.⇒x = (π -π/3) +2πk , k∈Z .ответ : 2π/3 +2πk , k∈Z .-------3.(2x² -1)^1/4 =(6x -4) ^ 1/4 ⇔ { 6x -4≥0 ;2x² -1 =6x-4.⇔{x≥2/3;2x² -6x+3 =0.{x≥2/3; [ x =(3 ±√3)/2 . ⇒ x =(3+√3)/2 .-------4. Lgsin2x =Lqsinx ⇔ {sin2x =sinx ; sinx > 0 .⇔{2sinx*cosx =sinx ; sinx>0.⇔ {cosx =1/2 ; sinx>0. ⇒ x =π/3 +2πn , n∈Z.-------5.x² -x +Lgcosx =1 +Lgcosx⇔{x² -x =1 ; cosx >0.⇔{ x² -x -1=0 ; cosx >0.⇔{x² -x -1=0 ; cosx >0 ⇔{ x=(1 ±√5)/2 ; cosx >0. ⇒ x=(1 -√5)/2.proizvodniyproizvodniyЛови-------------------------------------------------------------------------------------- в lgsinx=0 . в графике я не силен Новые вопросы в Алгебрасрочно помогите пожалуйста!!!!дам 50 балловочень срочно помогите пожалуйста!!!желательно с полными разборами:(!!!упростите выражение1/cos2a-tg2a+ctg2Терміново!! Будьласка допоможіть!!Знайдіть точки екстремуму та екстремуми функції f(x) =х^2+3х/х+4віддам усі свої балиПодайте число 16 у вигляді суми двох додатних доданків так, щоб їхдобуток був найбільшим.Знайдіть точки екстремуму та екстремуми функції f(x) =х^2+3х/х+47. Доведіть, що при будь-яких значеннях змінних x2 + y2 – 2(3x + y) + 10 ≥ 0розв'яжіть рівняння :ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Решите уравнения пожалуйста, ( кроме последнего) заранее спасибо...

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June 2022 1 28 Report

Решите уравнения пожалуйста, ( кроме последнего)
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Answers & Comments

proizvodniy 1.
√(2x+2) =1 -x⇔ {1 -x ≥ 0 ;2x +2 = (1-x)². ⇔{ x ≤ 1 ; x² -4x -1 =0.⇔
{ x ≤ 1 ; x= 2 ±√5.⇒ x= 2 -√5 .
-------
2.
√(8+10cosx) =2sinx . ⇔{ sinx ≥0 ; 8+10cosx =4sin²x. ⇔
{ sinx ≥0 ; 8+10cosx =4(1-cos²x) .⇔{ sinx ≥0 ; 2cos²x+5cosx +2=0.⇔
{ sinx ≥0 ; cos²x+(2+1/2)cosx +1=0.⇔{ sinx ≥0; [ cosx= -2(н.р.);cosx = -1/2. ⇔{ sinx ≥0; cosx = -1/2.⇒x = (π -π/3) +2πk , k∈Z .

ответ : 2π/3 +2πk , k∈Z .
-------
3.
(2x² -1)^1/4 =(6x -4) ^ 1/4 ⇔ { 6x -4≥0 ;2x² -1 =6x-4.⇔{x≥2/3;2x² -6x+3 =0.
{x≥2/3; [ x =(3 ±√3)/2 . ⇒ x =(3+√3)/2 .
-------
4. Lgsin2x =Lqsinx ⇔ {sin2x =sinx ; sinx > 0 .⇔{2sinx*cosx =sinx ; sinx>0.⇔
{cosx =1/2 ; sinx>0. ⇒ x =π/3 +2πn , n∈Z.
-------
5.
x² -x +Lgcosx =1 +Lgcosx⇔{x² -x =1 ; cosx >0.⇔{ x² -x -1=0 ; cosx >0.⇔
{x² -x -1=0 ; cosx >0 ⇔{ x=(1 ±√5)/2 ; cosx >0. ⇒ x=(1 -√5)/2.