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лоза10
@лоза10
July 2022
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решить уравнение
(х-2)(х²+2х+1)=4(х+1)
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lymovroma
(х-2)(х²+2х+1) = 4(х+1)
(х-2)(х+1)² - 4(х+1)=0
(х+1)((х-2)(х+1) - 4) = 0
х+1 = 0 (х-2)(х+1) - 4 = 0
х1 = -1 х² - х - 6 = 0
по т. Виетта
х2 = - 2, х3 = 3
0 votes
Thanks 2
crdd
(х-2)(х²+2х+1)=4(х+1)
(х-2)(х+1)²-4(х+1)=0
(х+1)((х-2)(х+1)-4)=0
(х+1)(х²-2х-2+х-4) =0
(х+1)(х²-х-6)=0
х+1=0
х=-1
или
х²-х-6=0
D=1+24=25
x=(1-5):2=-2
x=(1+5):2=3
ответ: -2;-1;3
1 votes
Thanks 2
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Answers & Comments
(х-2)(х+1)² - 4(х+1)=0
(х+1)((х-2)(х+1) - 4) = 0
х+1 = 0 (х-2)(х+1) - 4 = 0
х1 = -1 х² - х - 6 = 0
по т. Виетта
х2 = - 2, х3 = 3
(х-2)(х+1)²-4(х+1)=0
(х+1)((х-2)(х+1)-4)=0
(х+1)(х²-2х-2+х-4) =0
(х+1)(х²-х-6)=0
х+1=0
х=-1
или
х²-х-6=0
D=1+24=25
x=(1-5):2=-2
x=(1+5):2=3
ответ: -2;-1;3