题目内容
已知| 3 |
| x+y |
| 4 |
| y+z |
| 5 |
| z+x |
| x2+y2+z2 |
| xy+yz+zx |
分析:首先将已知
=
=
转化为
=
=
,并令
=
=
=a,通过解方程组用a分别表示x、y、z的值.再代入原式计算.
| 3 |
| x+y |
| 4 |
| y+z |
| 5 |
| z+x |
| x+y |
| 3 |
| y+z |
| 4 |
| x+z |
| 5 |
| x+y |
| 3 |
| y+z |
| 4 |
| x+z |
| 5 |
解答:解:∵
=
=
,
∴
?
?
=
=
∴令
=
=
=a
则有
由①+②+③得 x+y+z=6a ④
由④-①得 z=3a,
同理解得x=2a,y=a
∴
=
=
故答案为
.
| 3 |
| x+y |
| 4 |
| y+z |
| 5 |
| z+x |
∴
|
|
| x+y |
| 3 |
| y+z |
| 4 |
| x+z |
| 5 |
∴令
| x+y |
| 3 |
| y+z |
| 4 |
| x+z |
| 5 |
则有
|
由①+②+③得 x+y+z=6a ④
由④-①得 z=3a,
同理解得x=2a,y=a
∴
| x2+y2+z2 |
| xy+yz+zx |
| (22+12+32)a2 |
| (2×1+1×3+2×3)a2 |
| 14 |
| 11 |
故答案为
| 14 |
| 11 |
点评:本题考查分式的化简求值.首先将已知
=
=
转化为
=
=
,对于不确定的上述表达式,可令
=
=
=a,进而不难用a分别表示x、y、z的值,这是解决本题的关键.
| 3 |
| x+y |
| 4 |
| y+z |
| 5 |
| z+x |
| x+y |
| 3 |
| y+z |
| 4 |
| x+z |
| 5 |
| x+y |
| 3 |
| y+z |
| 4 |
| x+z |
| 5 |
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