решить уравнение 1 ; sqr(20/x + 1 ) - sqr(20/x - 1 ) = 6 2; sqr(20/x +1) + sqr(20/x - 1 ) = sqr(6). Created by bestseller61. algebra-ru

x - 1 ) = sqr(6)...

bestseller61 @bestseller61

August 2022 1 5 Report

решить уравнение 1 ; sqr(20/x + 1 ) - sqr(20/x - 1 ) = 6
2; sqr(20/x +1) + sqr(20/x - 1 ) = sqr(6)

Answers & Comments

oganesbagoyan

Verified answer

1). √(20/x +1) -√(20/x -1) =6 ;
√((x+20)/x) -√(-(x -20)/x) = 6 ;
ОДЗ :{ (x+20)/x ≥ 0 ;(x-20)/x ≤0. { x∈ (-∞; -20]  U (0;∞) ; x∈( 0;20]. ⇔ x∈( 0;20].
------------------------------------------------    или -------------------------------------
√(20/x +1) = 6 +√(20/x -1)  ;
(√(20/x +1))² = (6 +√(20/x -1))² ;
20/x +1 = 36 +12√(20/x -1) + 20/x -1
√(20/x -1) = -17/6 невозможно (√....... ≥ 0) ;
******************************************************************************8
2).  √(20/x +1) + √(20/x -1) = √6  ;
ОДЗ : x∈( 0;20]  смотри предыдущий пункт .
√(20/x +1) = √6 - √(20/x -1) ;
(√(20/x +1))² = (√6 - √(20/x -1))² ;²
20/x +1 = 6 -2√6*√(20/x -1) +20/x -1 ;
2√6*√(20/x -1) = 4 ;
√6*√(20/x -1) =2 ;
6*(20/x -1) = 4 ;
20/x -1 = 2/3 ;
20/x = 5/3 ;
4/x =1/3 ;
x=12 ∈ ОДЗ .
ответ:12  .
---------------------------------- проверка -------------------------------------
√(20/12 +1) + √(20/12 -1) = √16/6 +√4/6 =4/√6 + 2/√6 =6/√6=√6.

0 votes Thanks 0