найдите корень уравнения. Created by bobkovaleksand1. algebra-ru

найдите корень уравнения...

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ОДЗ
x ≠ 5 ; x ≠ - 5

3(x^2 - 25) = (x + 5)^2
3(x^2 - 25) - (x + 5)^2 = 0
3(x - 5)(x + 5) - (x + 5) = 0
(x + 5) (3(x - 5) - 1) = 0
(x + 5)*(3x - 15 - 1) = 0
(x + 5)*(3x - 16) = 0

x + 5 = 0
x = - 5 (не удовлетворяет ОДЗ)

3x - 16 = 0
3x = 16
x = 16/3 = 5 ц 1/3

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bobkovaleksand1 мне через ноз

oganesbagoyan

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Task/28137169
-------------------
1/(x² - 25) = 3 /(x+5)² ;
1/(x² - 5²) = 3/(x+5)² ; * * * x² - 5² = (x-5)(x+5) * * *
3/(x+5)² - 1/(x - 5) (x+5) = 0 ; * * * ОДЗ: x ≠ ± 5 * * *
( 3(x - 5) -(x+5) ) / (x - 5) (x+5)² =0 ;
2(x -10) / (x - 5) (x+5)² ; * * * 2 ≠0 * * *
x -10 =0 ;
x =10.

ответ: 10.

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