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darkis5
@darkis5
July 2022
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2,4 с решениями пожалуйста
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oganesbagoyan
Verified answer
2) ОДЗ:
x-3 ≠ 0 ⇒ x ≠ 3 . x∈(-∞;3) U (3 ;∞)
(x+4)(x-3)=0 ; x-3 ≠ 0 ОДЗ
⇒x+4 = 0
⇒
x= - 4 .
4) 5/(x+3) -2x/(x-3) =36/(9-x
²) ; ОДЗ ≠ (+/-)3.
2x/(x-3) -5/(x+3) = 36/(x² -9);
2x(x+3) -5(x-3) =36;
2x² +x -15 =0;
x₁
= (-1 -sqrt(1² - 4*2*(-15))/(2*2) = (1-sqrt121)/4=(1-11)/4
=
-2,5;
x₂
= (-1 + sqrt(1² - 4*2*(-15))/(2*2) = (1+sqrt121)/4=(1+11)/4
= 3
.
1 votes
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007gaglik
Номер 4. Надеюсь понятно
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Answers & Comments
Verified answer
2) ОДЗ: x-3 ≠ 0 ⇒ x ≠ 3 . x∈(-∞;3) U (3 ;∞)(x+4)(x-3)=0 ; x-3 ≠ 0 ОДЗ
⇒x+4 = 0 ⇒ x= - 4 .
4) 5/(x+3) -2x/(x-3) =36/(9-x²) ; ОДЗ ≠ (+/-)3.
2x/(x-3) -5/(x+3) = 36/(x² -9);
2x(x+3) -5(x-3) =36;
2x² +x -15 =0;
x₁= (-1 -sqrt(1² - 4*2*(-15))/(2*2) = (1-sqrt121)/4=(1-11)/4 = -2,5;
x₂ = (-1 + sqrt(1² - 4*2*(-15))/(2*2) = (1+sqrt121)/4=(1+11)/4 = 3.