题目内容
(1)已知b2=ac,求| a2b2c2 |
| a3+b3+c3 |
| 1 |
| a3 |
| 1 |
| b3 |
| 1 |
| c3 |
(2)已知x、y、z满足
| x |
| y+z |
| y |
| z+x |
| z |
| x+y |
| x2 |
| y+z |
| y2 |
| z+x |
| z2 |
| x+y |
分析:(1)先把分式化简,再代入求值即可;
(2)由已知可得
=1-
-
,则
=x-
-
,同理求得所求代数式中的后两个式子的表达式,相加并化简即可.
(2)由已知可得
| x |
| y+z |
| y |
| z+x |
| z |
| x+y |
| x2 |
| y+z |
| xy |
| z+x |
| yz |
| x+y |
解答:解:(1)原式=
•
=
•
=
=
,
∵b2=ac,
∴原式=1;
(2)由已知可得
=1-
-
,
则
=x-
-
①,
同理
=y-
-
②,
=z-
-
③,
①+②+③得
+
+
=(x+y+z)-
-
-
=x+y+z-y-x-z=0.
| a2b2c2 |
| a3+b3+c3 |
| b3c3+a3c3+a3b3 |
| a3b3c3 |
| 1 |
| a3+b3+c3 |
| b3c3+a3b3+a3c3 |
| abc |
| b3(a3+b3+c3) |
| (a3+b3+c3)abc |
| b2 |
| ac |
∵b2=ac,
∴原式=1;
(2)由已知可得
| x |
| y+z |
| y |
| z+x |
| z |
| x+y |
则
| x2 |
| y+z |
| xy |
| z+x |
| xz |
| x+y |
同理
| y2 |
| z+x |
| xy |
| y+z |
| yz |
| x+y |
| z2 |
| x+y |
| xz |
| y+z |
| yz |
| z+x |
①+②+③得
| x2 |
| y+z |
| y2 |
| z+x |
| z2 |
| x+y |
| xy+yz |
| z+x |
| xz+xy |
| y+z |
| xz+yz |
| x+y |
点评:此题考查了分式的化简求值,要特别注意观察已知条件和所求代数式的关系,再进行化简.此题难度较大.
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