题目内容
已知x=5+2
,y=5-2
,求下列各式的值:
(1)x+y
(2)xy
(3)2x2+4xy+2y2
(4)
.
| 6 |
| 6 |
(1)x+y
(2)xy
(3)2x2+4xy+2y2
(4)
| x2-80xy+y2 |
考点:二次根式的化简求值
专题:
分析:(1)(2)直接代入求得答案即可;
(3)(4)先分解因式,再把(1)(2)的结果代入求得答案即可.
(3)(4)先分解因式,再把(1)(2)的结果代入求得答案即可.
解答:解:∵x=5+2
,y=5-2
,
∴(1)x+y
=5+2
+5-2
=10;
(2)xy
=(5+2
)(5-2
)
=25-24
=1;
(3)2x2+4xy+2y2
=2(x+y)2
=200;
(4)
=
=
=3
.
| 6 |
| 6 |
∴(1)x+y
=5+2
| 6 |
| 6 |
=10;
(2)xy
=(5+2
| 6 |
| 6 |
=25-24
=1;
(3)2x2+4xy+2y2
=2(x+y)2
=200;
(4)
| x2-80xy+y2 |
=
| (x+y)2-82xy |
=
| 100-82 |
=3
| 2 |
点评:此题考查二次根式的化简求值,注意整体代入思想的渗透.
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