Помогите пожалуйчта решить уравнение номер 3*). Created by КрОлИкБаНнИ. algebra-ru

Помогите пожалуйчта решить уравнение номер 3*)...

mmb1

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(x2-7x+3)^2+10(x2-7x+3)+21=0
x2-7x+3=t
t2+10t+21=0
t12=(-10+-корень(100-84))/2=(-10+-4)/2=-7 -3
x2-7x+3=-3
x2-7x+6=0
x12=(7+-корень(49-24))/2=(7+-5)/2=6 1
x2-7x+3=-7
x2-7x+10=0
x34=(7+-коренб(49-40))/2=(7+-3)/2=5 2
x= 1 2 5 6

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Света19999

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(x^2 - 7x + 3)^2 + 10(x^2 - 7x + 3) + 21 = 0
Пусть x^2 - 7x + 3 = t
t^2 + 10t + 21 = 0
D/4 = 25 - 21 = 4
t1 = -5 + 2 = -3
t2 = -5 - 2 = -7
Выполним обратную замену:
x^2 - 7x + 3 = -3                    или                        x^2 - 7x + 3 = -7
x^2 - 7x + 6 = 0                                                  x^2 - 7x + 10 = 0
D = 49 - 24 = 25                                                 D = 49 - 40 = 9
x1 = (7+5)/2 = 6                                                 x3 = (7+3)/2 = 5
x2 = (7-5)/2 = 1                                                  x4 = (7-3)/2 = 2
Ответ: 1;2;5;6.

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