题目内容

a
x
=
b
y
=
c
z
,求证:
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:
a
x
=
b
y
=
c
z
=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.
解答:解:设
a
x
=
b
y
=
c
z
=t,则a=xt,y=bt,z=ct,
a3
x2
+
b3
y2
+
c3
z2
=
x3t3
x2
+
y3t3
y2
+
z3t3
z2
=xt3+yt3+zt3=t3(x+y+z),
(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
=t3(x+y+z).
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
a
x
=
b
y
=
c
z
=t,则a=xt,y=bt,z=ct.
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