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160475o
@160475o
July 2022
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Решите уравнение(оно должно быть квадратным)
(Х-5)^2+3 (х-5)-10=0
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(x-5)(x-5)+3*(x-5)-10=0
(x-5)=t
t^2+3t-10=0
D = 9-4*1*-10=49
t_1,t_2= -3+-7/2
t_1 = 2, t_2 = - 5
(x-5)=2
x=7
(x-5)=-5
x=0
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Answers & Comments
(x-5)=t
t^2+3t-10=0
D = 9-4*1*-10=49
t_1,t_2= -3+-7/2
t_1 = 2, t_2 = - 5
(x-5)=2
x=7
(x-5)=-5
x=0