题目内容
如果a2+b2+c2=1,a+b+c=0,化简或求值:a2(
+
)+b2(
+
)+c2(
+
).
| 1 |
| b |
| 1 |
| c |
| 1 |
| a |
| 1 |
| c |
| 1 |
| a |
| 1 |
| b |
考点:分式的化简求值
专题:计算题
分析:先利用单项式乘以多项式的法则展开并整理,再把a+b+c=0两边平方求出ab+bc+ac的值,然后代入计算即可得解.
解答:解:a2(
+
)+b2(
+
)+c2(
+
),
=
+
+
+
+
+
=(
+
)+(
+
)+(
+
)
=
+
+
=
+
+
=
+
+
-a-b-c
=
-(a+b+c)
=
,
∵a+b+c=0,
∴(a+b+c)2=0,
∴a2+b2+c2+2ab+2bc+2ac=0,
∴ab+bc+ac=-
,
∴原式=-
.
| 1 |
| b |
| 1 |
| c |
| 1 |
| a |
| 1 |
| c |
| 1 |
| a |
| 1 |
| b |
=
| a2 |
| b |
| a2 |
| c |
| b2 |
| a |
| b2 |
| c |
| c2 |
| a |
| c2 |
| b |
=(
| a2 |
| b |
| c2 |
| b |
| a2 |
| c |
| b2 |
| c |
| b2 |
| a |
| c2 |
| a |
=
| a2+c2 |
| b |
| a2+b2 |
| c |
| b2+c2 |
| a |
=
| 1-b2 |
| b |
| 1-c2 |
| c |
| 1-a2 |
| a |
=
| 1 |
| a |
| 1 |
| b |
| 1 |
| c |
=
| ab+bc+ac |
| abc |
=
| ab+bc+ac |
| abc |
∵a+b+c=0,
∴(a+b+c)2=0,
∴a2+b2+c2+2ab+2bc+2ac=0,
∴ab+bc+ac=-
| 1 |
| 2 |
∴原式=-
| 1 |
| 2abc |
点评:本题考查了分式的化简求值,主要利用了完全平方公式和分式的加减混合运算,整理并求出ab+bc+ac的值是解题的关键.
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