已知：a+b+c=0 1/(a+1) +1/(b+2) +1/(c+3)=0 求(a+1)^2+(b+2)^2+(c+3)^2=? 注：^2为平方
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a+b+c=0
得：(a+1)+(b+2)+(c+3)=6
[(a+1)+(b+2)+(c+3)]^2=(a+1)^2+(b+2)^2+(c+3)^2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36
1/(a+1) +1/(b+2) +1/(c+3)=0
得：[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]/[(a+1)(b+2)(c+3)]=0
[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=0
所以(a+1)^2+(b+2)^2+(c+3)^2＝36-2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36-0＝36
得：(a+1)+(b+2)+(c+3)=6
[(a+1)+(b+2)+(c+3)]^2=(a+1)^2+(b+2)^2+(c+3)^2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36
1/(a+1) +1/(b+2) +1/(c+3)=0
得：[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]/[(a+1)(b+2)(c+3)]=0
[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=0
所以(a+1)^2+(b+2)^2+(c+3)^2＝36-2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36-0＝36
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