题目内容
如果a+b+c=0,
+
+
=0,求(a+1)2+(b+2)2+(c+3)2的值.
| 1 |
| a+1 |
| 1 |
| b+2 |
| 1 |
| c+3 |
分析:利用完全平方公式(a+b+c)2=a2+b2+c2+2ab+2bc+2ac,得a2+b2+c2=(a+b+c)2-2ab-2bc-2ac,仿照这个公式即可.
解答:解:由
+
+
=0,去分母,得
(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0,
而(a+1)2+(b+2)2+(c+3)2
=[(a+1)+(b+2)+(c+3)]2-2[(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)]
=(a+b+c+6)2
=(0+6)2
=36.
| 1 |
| a+1 |
| 1 |
| b+2 |
| 1 |
| c+3 |
(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0,
而(a+1)2+(b+2)2+(c+3)2
=[(a+1)+(b+2)+(c+3)]2-2[(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)]
=(a+b+c+6)2
=(0+6)2
=36.
点评:本题主要考查完全平方公式,熟记公式的几个变形公式对解题大有帮助.
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