问题 解答题
已知:
x
=
a
+
1
a
(0<a<1),求代数式
x2+x-6
x
÷
x+3
x2-2x
-
x-2+
x2-4x
x-2-
x2-4x
的值.
答案

x
=
a
+
1
a

∴x=a+

1
a
+2,

x-2=a+

1
a
,(x-2)2=(a+
1
a
2

即:x2-4x=a2+

1
a2
-2=(a-
1
a
2

∴原式=

(x+3)(x-2)
x
x(x-2)
x+3
x-2+
x2-4x
x-2-
x2-4x
=(x-2)2-
a+
1
a
+
(a-
1
a
)
2
a+
1
a
-
(a-
1
a
)
2
=(a+
1
a
2-
a+
1
a
+
(a-
1
a
)
2
a+
1
a
-
(a-
1
a
)
2

∵0<a<1,∴a-

1
a
<0,

∴原式=a2+

1
a2
+2-
a+
1
a
-a+
1
a
a+
1
a
+a-
1
a

=a2+

1
a2
+2-
1
a2

=a2+2.

问答题 简答题
判断题