题目内容
解方程组:
.
|
分析:首先令
=t(t≥0),把x+y+z=t2-5,代入①解出t的值,然后求出x+y+z=11,再由式②及等比定理得:
=
=
=
=
,于是求出x、y和z的值.
| x+y+z+5 |
| x |
| 3 |
| y |
| 4 |
| z |
| 5 |
| x+y+z |
| 12 |
| 11 |
| 12 |
解答:解:令
=t(t≥0),
则x+y+z=t2-5,代入①得:
2(t2-5)-5t=2,
即2t2-5t-12=0,
解得t=4或t=-
(舍去),
由
=4得:
x+y+z=11,
由式②及等比定理得:
=
=
=
=
,
解得:x=
,y=
,z=
.
| x+y+z+5 |
则x+y+z=t2-5,代入①得:
2(t2-5)-5t=2,
即2t2-5t-12=0,
解得t=4或t=-
| 3 |
| 2 |
由
| x+y+z+5 |
x+y+z=11,
由式②及等比定理得:
| x |
| 3 |
| y |
| 4 |
| z |
| 5 |
| x+y+z |
| 12 |
| 11 |
| 12 |
解得:x=
| 11 |
| 4 |
| 11 |
| 3 |
| 55 |
| 12 |
点评:本题主要考查无理方程的知识点,解答本题的关键是熟练运用等比定理,此题难度不是很大.
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