若a+b+c=0,1/a+1+1/b+2+1/c+3=0,那么(a+1)2+(b+2)2+(c+3)2=_.

问题描述:

若a+b+c=0,

1
a+1
+
1
b+2
+
1
c+3
=0,那么(a+1)2+(b+2)2+(c+3)2=______.

∵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.