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Ilyasssssss
@Ilyasssssss
July 2022
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Найдите сумму корней уравнения
(х-7)(х-5)(х-3)(х-1)=40
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sedinalana
Verified answer
[(x-7)(x-1)}8[(x-5)(x-3)]-40=0
(x²-8x+7)*(x²-8x+15)-40=0
x²-8x+7=a
a*(a+8)-40=0
a²+8a-40=0
D=64+160=224
a1=(-8-4√14)/2=-4-2√14 U a2=-4+2√14
x²-8x+7=-4-2√14
x²-8x+11+2√14=0
D=64-44-8√14=20-8√14<0 нет решения
x²-8x+7=-4+2√14
x²-8x+11-2√14=0
D=64-44+8√14=20+8√14
x1=
x2=
2 votes
Thanks 1
Ilyasssssss
ответ 8,
Ilyasssssss
8?
mmb1
если посчитать среднее (7+5+3+1)/4=4 и сделать замену х-4=t то получится (t+3)(t+1)(t-1)(t-3)=40 (t^2-9)(t^2-1)=40 t^4-10t^2-31=0
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Answers & Comments
Verified answer
[(x-7)(x-1)}8[(x-5)(x-3)]-40=0(x²-8x+7)*(x²-8x+15)-40=0
x²-8x+7=a
a*(a+8)-40=0
a²+8a-40=0
D=64+160=224
a1=(-8-4√14)/2=-4-2√14 U a2=-4+2√14
x²-8x+7=-4-2√14
x²-8x+11+2√14=0
D=64-44-8√14=20-8√14<0 нет решения
x²-8x+7=-4+2√14
x²-8x+11-2√14=0
D=64-44+8√14=20+8√14
x1=
x2=