A+B+C=2√(A-1)+4√(B+1)+6√(C-2)-12
[(A-1)-2√(A-1)+1]+[(B+1)-4√(B+1)+4]+[C-2+9-6√(C-2)]=0
[√(A-1)-1]^2+[√(B+1)-2]^2+[√(C-2)-3]^2=0
A=2
B=3
C=11
代入
122
2A+2B-4√(A-1)-8√(B-2)+C-6√(C-3)+10=0
[2(A-1)-4√(A-1)+2]+[2(B-2)-8√(B-2)+8]+[C-3-6√(C-3)+9]=0
√(A-1)=1
A=2
√(B-2)=2
B=6
√(C-3)=3
C=12
求A+B+C=20
3.[1+1/n-1/(n+1)]^2
=[1+1/n(n+1)]^2
=1+2/n(n+1)+[1/n(n+1)]^2
=1+2/n(n+1)+[1/n-1/(n+1)]^2(平方展开)
=1+1/n^2+1/(n+1)^2
所以S＝(1+1/1-1/2)+(1+1/2-1/3)+……+(1+1/2003-1/2004)
=2003-1/2004
所以S的整数部分是2002
正数abc满足a+c=2b
求证
1/(√a+√b)+1/(√b+√c)=2/(√a+√c)
a=b+t
c=b-t
1/(√a+√b)=(√a-√b)/(√a+√b)(√a-√b)=(√a-√b)/t
1/(√b+√c)=(√b-√c)/(√b+√c)(√b-√c)=(√b-√c)/t
2/(√a+√c)=2(√a-√c)/(√a+√c)(√a-√c)=2(√a-√c)/2t=(√a-√c)/t
证完啦