Решите Уравнение прошу : 4 номер. Created by frankmiki. algebra-ru

Решите Уравнение прошу : 4 номер...

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х²+х-2/х-3 = 2х+4/х(х-3)
х³+х²-2х=2х+4
х³+х²-4х-4=0
(х³+х²)-(4х+4)=0
х²(х+1)-4(х+1)=0
(х²-4)(х+1)=0
х²-4=0 или х+1=0
х1=2
х2=-2
х3=-1

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oganesbagoyan

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(x² +x -2) /(x-3) =(2x+4) / (x² -3x) ;
(x+2)(x-1) /(x-3) = 2(x+2) / x(x-3) ;
ОДЗ : x ≠0 ; x ≠3. * * * x∈(-∞ ; 0) ∪ (0;3) ∪ (3; ∞)
x₀ = -2 корень .
если x ≠ -2 , то
(x-1)x = 2 ;
x² - x -2 =0 ;
x₁ = (1 +3)/2 =2 ;
x ₂ =(1 -3) /2 = -1 .

ответ : { -2 ; -1 ; 2} .

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