Question

如果a+b+c=0,a+1分之1+b+1分之1+c+3分之1=0,求(a+1)平方+(b+2)平方+(c+3)平方的值

Answer 1

a+b+c=0
(a+1)+(b+2)+(c+3)=6平方
(a+1)^2+(b+2)^2+(c+3)^2+2(a+1)(b+2)+2(a+1)(c+3)+2(b+2)(c+3)=36
(a+1)^2+(b+2)^2+(c+3)^2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36.............1

1/(a+1)+1/（b+2）+1/(c+3)=0 去分母
(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)=0...................2

将2式代入1式得
(a+1)^2+(b+2)^2+(c+3)^2=36

Answer 2

∵a+b+c=0
∴a+1+b+2+c+3=1+2+3=6
令x=a+1，y=b+2，z=c+3
那么由题意得：x+y+z=6
①，1/x+1/y+1/z=0
②
由①得：(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=36
③
由②得：(xy+xz+yz)/(xyz)=0，所以xy+xz+yz=0
④
由③④式得：x^2+y^2+z^2=36
即：(a+1)平方+(b+2)平方+(c+3)平方=36