若a+b+c=0,1a+1+1b+2+1c+3=0

学习 时间:2026-03-30 17:43:10 阅读:8402
若a+b+c=0,1a+1+1b+2+1c+3=0

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自然的金鱼

含蓄的宝马

2026-03-30 17:43:10

∵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1a+1+1b+2+1c+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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  • 兴奋的狗
    回复
    2026-03-30 17:43:10

    ∵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1a+1+1b+2+1c+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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