若a+b+c=0,1a+1+1b+2+1c+3=0
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若a+b+c=0,
+
+
=0
1 |
a+1 |
1 |
b+2 |
1 |
c+3 |
∵a+b+c=0,
∴(a+1)+(b+2)+(c+3)=6,
两边平方得(a+1)2+(b+2)2+(c+3)2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36,
又由
1
a+1+
1
b+2+
1
c+3=0去分母,得
(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0,
∴(a+1)2+(b+2)2+(c+3)2=36.
故答案为:36.
∴(a+1)+(b+2)+(c+3)=6,
两边平方得(a+1)2+(b+2)2+(c+3)2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36,
又由
1
a+1+
1
b+2+
1
c+3=0去分母,得
(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0,
∴(a+1)2+(b+2)2+(c+3)2=36.
故答案为:36.
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