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AnnaLange
@AnnaLange
July 2022
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Решите уравнения :
1) (х+1)^4+(х+1)²-6=0
2) (6/х²-4х+3)-(13-7х/1-х)=3/х-3
^ - в степени
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Galina2013
1) (x + 1)^2 = t
t^2 + t - 6 = 0
D = 1 + 24 = 25
t1 = 2
t2 = -3 (не подходит, так как квадрат числа всегда > 0)
(x + 1)^2 = 2
x^2 + 2t - 1 = 0
D = 4 - 4 = 0
x = -1
1 votes
Thanks 1
AnnaLange
Спасибо , сможешь 2 решить ?
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Answers & Comments
t^2 + t - 6 = 0
D = 1 + 24 = 25
t1 = 2
t2 = -3 (не подходит, так как квадрат числа всегда > 0)
(x + 1)^2 = 2
x^2 + 2t - 1 = 0
D = 4 - 4 = 0
x = -1