问题 解答题
①已知x=2-
3
,y=2+
3
,求:x2+xy+y2的值.
②已知x=
2
+1
,求x+1-
x2
x-1
的值.
答案

①x2+xy+y2=(x2+2xy+y2)-xy

=(x+y)2-xy,

当x=2-

3
,y=2+
3
时,

原式=(2-

3
+2+
3
2-(2-
3
)(2+
3

=16-1=15;

②x+1-

(x+1)(x-1)-x2
x-1
=
x2-1-x2
x-1
=
-1
x-1

当x=

2
+1时,

-1
x-1
=
-1
2
+1-1
=-
2
2

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