若m,n是两个不相等的实数,且满足㎡-2m=1,n^2-2n=1,求代数式2㎡+4n^2-4n+1999的值
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![若m,n是两个不相等的实数,且满足㎡-2m=1,n^2-2n=1,求代数式2㎡+4n^2-4n+1999的值](/uploads/image/z/1811499-51-9.jpg?t=%E8%8B%A5m%2Cn%E6%98%AF%E4%B8%A4%E4%B8%AA%E4%B8%8D%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AE%9E%E6%95%B0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E3%8E%A1-2m%3D1%2Cn%5E2-2n%3D1%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F2%E3%8E%A1%2B4n%5E2-4n%2B1999%E7%9A%84%E5%80%BC)
若m,n是两个不相等的实数,且满足㎡-2m=1,n^2-2n=1,求代数式2㎡+4n^2-4n+1999的值
若m,n是两个不相等的实数,且满足㎡-2m=1,n^2-2n=1,求代数式2㎡+4n^2-4n+1999的值
若m,n是两个不相等的实数,且满足㎡-2m=1,n^2-2n=1,求代数式2㎡+4n^2-4n+1999的值
解:由已知得,m和n是方程x^2-2x-1=0的两个根.
根据根与系数的关系,m+n=2,
m^2=2m+1. n^2=2n+1
原式=2(2m+1)+4(2n+1)-4n+1999
=4m+2+8n+4-4n+1999
=4(m+n)+2005
=4×2+2005
=2013
m^2-2m=1,n^2-2n=1
可以解得
m=n=根号2+1
m^2=1+2m
2㎡+4n^2-4n+1999=6m^2-4m+1999=6(1+2m)-4m+1999
=6+8m+1999
=2005+8(根号2+1)
=2013+8*根号2
m^2-2m=1,n^2-2n=1
可以解得
m=根号2+1 ,n=-根号2+1;或m=-根号2+1 ,n=根号2+1
2㎡+4n^2-4n+1999=㎡+㎡+n^2+3n^2-4n+1999=[㎡+(1+2m)]+[n^2+3(1+2n)-4n]+1999=(m+1)^2+[(n+1)^2+2]+1999
无论是m=根号2+1 ,n=-根号2+1还是m=-根号...
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m^2-2m=1,n^2-2n=1
可以解得
m=根号2+1 ,n=-根号2+1;或m=-根号2+1 ,n=根号2+1
2㎡+4n^2-4n+1999=㎡+㎡+n^2+3n^2-4n+1999=[㎡+(1+2m)]+[n^2+3(1+2n)-4n]+1999=(m+1)^2+[(n+1)^2+2]+1999
无论是m=根号2+1 ,n=-根号2+1还是m=-根号2+1 ,n=根号2+1,
上式都等于2+4根号2+4+2-4根号2+4+2+1999=2011
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n^2啥意思?
如果题目没有写错的话,m和n其实是x^2-2x=1的两个解。用根与系数关系,可知:
M+N = -b/a = 2
MN = c/a = -1
M^2+N^2 = (M+N)^2 - 2MN = 6
那个代数式就可以写成 2(M^2 + N^2) + 2(N^2 - 2N) + 1999
也就等于2*6+2*1+1999=2013