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yugolovin
@yugolovin
July 2022
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Решить уравнение
[tex]\sqrt{8x+1}+\sqrt{25x+6}=\sqrt{3x+19}[/tex]
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Дмитрий1425
Task/25502422
----------------------
Решить уравнение
√(8x +1) + √(25x +6) =
√(3x +19) .
ОДЗ:
{ 8x +1 ≥0 ; 25x +6 ≥0 ; 3x +19 ≥0 .⇒
x
≥ - 1/8 = - 0,125.
------- возведем в квадрат -------
8x +1 + 2√(8x +1)
(25x +6) + 25x +6 = 3x +19 ;
√(8x +1)
(25x +6) = -3(5x -2) ;
* * *
5x -2 ≤ 0 , т.е x ≤ 0,4 таким образом :
x ∈ [- 0,125 ,0,4] * * *
(8x +1)
(25x +6) = 9(5x -2 )
² ;
200x² + 73x +6 =225x² -180x +36 ;
25x² -253x +30 =0 ; * * * x² - (250+3)/25+30/25 =0 * * *
x² - (10 +3/25)+10*(3/25) =0 ;
x₁ =10 ∉
[- 0,125 ,0,4]
x₂ = 3/25 = 0, 12 .
ответ :
0, 12 .
3 votes
Thanks 1
yugolovin
Чтобы сделать "плюс-минус", надо воспользоваться командой \pm
yugolovin
Все правильно, спасибо
oganesbagoyan
Verified answer
1 votes
Thanks 0
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Answers & Comments
----------------------
Решить уравнение
√(8x +1) + √(25x +6) = √(3x +19) .
ОДЗ: { 8x +1 ≥0 ; 25x +6 ≥0 ; 3x +19 ≥0 .⇒ x ≥ - 1/8 = - 0,125.
------- возведем в квадрат -------
8x +1 + 2√(8x +1)(25x +6) + 25x +6 = 3x +19 ;
√(8x +1)(25x +6) = -3(5x -2) ;
* * * 5x -2 ≤ 0 , т.е x ≤ 0,4 таким образом : x ∈ [- 0,125 ,0,4] * * *
(8x +1)(25x +6) = 9(5x -2 )² ;
200x² + 73x +6 =225x² -180x +36 ;
25x² -253x +30 =0 ; * * * x² - (250+3)/25+30/25 =0 * * *
x² - (10 +3/25)+10*(3/25) =0 ;
x₁ =10 ∉ [- 0,125 ,0,4]
x₂ = 3/25 = 0, 12 .
ответ : 0, 12 .
Verified answer