题目内容
已知x=
+
,y=
-
,求下列各式的值:
(1)xy2+x2y;
(2)x2-xy+y2.
| 5 |
| 6 |
| 5 |
| 6 |
(1)xy2+x2y;
(2)x2-xy+y2.
分析:先求得x+y=2
,x-y=2
;xy=-1.
(1)把所求的代数式转化为xy(x+y),然后将其代入求值即可;
(2)把所求的代数式转化为(x-y)2+xy,然后将其代入求值即可.
| 5 |
| 6 |
(1)把所求的代数式转化为xy(x+y),然后将其代入求值即可;
(2)把所求的代数式转化为(x-y)2+xy,然后将其代入求值即可.
解答:
解:∵x=
+
,y=
-
,
∴x+y=2
,x-y=2
;xy=-1.
(1)xy2+x2y=xy(x+y),=-1×2
=-2
;
(2)x2-xy+y2=(x-y)2+xy=(2
)2-1=23.
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| 6 |
| 5 |
| 6 |
∴x+y=2
| 5 |
| 6 |
(1)xy2+x2y=xy(x+y),=-1×2
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(2)x2-xy+y2=(x-y)2+xy=(2
| 6 |
点评:本题考查了二次根式的化简求值.二次根式的化简求值,一定要先化简再代入求值.
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