[tex]3. (x+6)^4+(x+4)^4=82 4. (x+3)(x^2-4)(x+7)+96=0[/tex]. Created by Azbox. algebra-ru

tex]...

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[tex]3. (x+6)^4+(x+4)^4=82

4. (x+3)(x^2-4)(x+7)+96=0[/tex]

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oganesbagoyan

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3.
(x+6)⁴ +(x+4)⁴ =82 ;
(x+5+1)⁴  +(x+5 -1)⁴ =82 ; * * * t =x+5 * * *
(t+1)⁴  +(t -1)⁴ =82 ;
(t⁴ +4t³ +6t² +4t +1) +(t⁴ -4t³ +6t² -4t +1) =82 ;
2t⁴ +12t² -80 =0 ;
t⁴ +6t² -40 =0 ;
* * *   биквадратное уравнение z = t² ≥ 0 ; z² +6z -40 =0 * * *
или
t⁴ +10t² -4t² - 40  =0  ;
t² (t² +10) -4(t² +10) =0 ;
(t² +10)(t² -4) = 0 ;
[ t² = -10 (нет действительных корней) ; t² =4.
⇒(x+5)² =4⇔x+5 = ±2 ⇔  [ x+5 = -2 ;  x+5 = 2. ⇔[ x = -7 ;  x = -3.
ответ : {-7 ; -3}.
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4.
(x+3)(x² -4)(x+7) +96=0 ;
(x+3)(x+2)(x-2)(x+7) +96=0 ;
(x² +5x+6)(x² +5x -14) +96 =0 ;
* * *замена t= x² +5x+6 ⇒ x² +5x -14=(x² +5x +6) -20 = t -20* * *
---
t(t -20) +96 =0 ;
t² -20t +96 =0 ;
[t =8 ; [t =12 .
а)  x² +5x+6 =8 ⇔x² +5x-2 =0 ⇒x₁ =(-5 - √33)/2 ; x₂ = (-5+√33)/2;
б) x² +5x+6 =12 ⇔x² +5x-6 =0  ⇒x₃ = -6 ; x₄=1.

ответ : {(-5 - √33)/2 ; (-5+√33)/2; ; -6; 1}.

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