题目内容
计算:(x-2-y-2)÷(x-1-y-1)(结果不含负整数指数幂).
分析:方法一:根据负整数指数次幂等于正整数指数次幂的倒数转化为分式,再根据分式的加减运算以及除法运算进行计算即可得解;
方法二:先把被除数利用平方差公式分解因式,然后约分,再根据负整数指数次幂等于正整数指数次幂的倒数进行计算即可得解.
方法二:先把被除数利用平方差公式分解因式,然后约分,再根据负整数指数次幂等于正整数指数次幂的倒数进行计算即可得解.
解答:解:方法一:(x-2-y-2)÷(x-1-y-1),
=(
-
)÷(
-
),
=
÷
,
=
•
,
=
;
方法二:(x-2-y-2)÷(x-1-y-1),
=(x-1-y-1)(x-1+y-1)÷(x-1-y-1),
=x-1+y-1,
=
+
,
=
.
=(
| 1 |
| x2 |
| 1 |
| y2 |
| 1 |
| x |
| 1 |
| y |
=
| y2-x2 |
| x2y2 |
| y-x |
| xy |
=
| (y-x)(y+x) |
| x2y2 |
| xy |
| y-x |
=
| x+y |
| xy |
方法二:(x-2-y-2)÷(x-1-y-1),
=(x-1-y-1)(x-1+y-1)÷(x-1-y-1),
=x-1+y-1,
=
| 1 |
| x |
| 1 |
| y |
=
| x+y |
| xy |
点评:本题主要考查了负整数指数次幂等于正整数指数次幂的倒数的性质,分式的混合运算,熟练掌握负整数指数幂的性质是解题的关键.
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