题目内容
若
=
=
,求证:
+
+
=
.
| a |
| x |
| b |
| y |
| c |
| z |
| a3 |
| x2 |
| b3 |
| y2 |
| c3 |
| z2 |
| (a+b+c)3 |
| (x+y+z)2 |
考点:分式的混合运算,比例的性质
专题:证明题
分析:设
=
=
=t,则a=xt,y=bt,z=ct,再分别代入式子的左右两边计算后即可得证.
| a |
| x |
| b |
| y |
| c |
| z |
解答:解:设
=
=
=t,则a=xt,y=bt,z=ct,
+
+
=
+
+
=xt3+yt3+zt3=t3(x+y+z),
=
=
=t3(x+y+z).
所以
+
+
=
.
| a |
| x |
| b |
| y |
| c |
| z |
| a3 |
| x2 |
| b3 |
| y2 |
| c3 |
| z2 |
| x3t3 |
| x2 |
| y3t3 |
| y2 |
| z3t3 |
| z2 |
| (a+b+c)3 |
| (x+y+z)2 |
| (xt+yt+zt)3 |
| (x+y+z)2 |
| t3(x+y+z)3 |
| (x+y+z)2 |
所以
| a3 |
| x2 |
| b3 |
| y2 |
| c3 |
| z2 |
| (a+b+c)3 |
| (x+y+z)2 |
点评:本题考查了分式的混合运算以及比列的性质,关键是设
=
=
=t,则a=xt,y=bt,z=ct.
| a |
| x |
| b |
| y |
| c |
| z |
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