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arina1536
@arina1536
August 2022
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Пожалуйста помогите!
Решите пример не в лоб - 2004х^2 - 2003х - 1 = 0
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mikael2
Verified answer
Попробуем в общем виде (n+1)x²-nx-1 D=n²+4(n+1)=n²+4n+4=(n+2)²
√D=n+2 >0 x1=1/2*(n+1)[n-n-2]=-2(n+1)/2=-(n+1)
x2=1/2*(n+1)(n+n+2)=(n+1)(n+1)=(n+1)²
n=2003 x1=-2004 x2=2003²=4 012 009
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Answers & Comments
Verified answer
Попробуем в общем виде (n+1)x²-nx-1 D=n²+4(n+1)=n²+4n+4=(n+2)²√D=n+2 >0 x1=1/2*(n+1)[n-n-2]=-2(n+1)/2=-(n+1)
x2=1/2*(n+1)(n+n+2)=(n+1)(n+1)=(n+1)²
n=2003 x1=-2004 x2=2003²=4 012 009