如果m,n是两个不相等的实数,且满足m2-2m=1,n2-2n=1,那么代数式2m2+4n2-4n+1999=______.
人气:265 ℃ 时间:2019-08-18 13:25:45
解答
由题意可知:m,n是两个不相等的实数,且满足m2-2m=1,n2-2n=1,所以m,n是x2-2x-1=0两个不相等的实数根,则根据根与系数的关系可知:m+n=2,又m2=2m+1,n2=2n+1,则2m2+4n2-4n+1999=2(2m+1)+4(2n+1)-4n+1999=4m...
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