题目内容

(1)化简
3a
9
2
a-3
÷
3a-7
3a13

(2)已知x+y=12,xy=9,且0<x<y,求
x
1
2
-y
1
2
x
1
2
+y
1
2
的值.
(1)原式=(a
9
2
a-
3
2
)
1
3
÷(a-
7
3
a
13
3
)
1
2
=(a3)
1
3
÷(a2)
1
2
=a÷a=1;
(2)号
x
1
2
-y
1
2
x
1
2
+y
1
2
=
(
x
-
y
)2
(
x
+
y
)(
x
-
y
)
=
x+y-2
xy
x-y

∵x+y=12,xy=9,且0<x<y,
x-y=-
(x-y)2
=-
(x+y)2-4xy
=-
122-4×9
=-6
3

x
1
2
-y
1
2
x
1
2
+y
1
2
=
x+y-2
xy
x-y
=
12-2×3
-6
3
=-
3
3
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