31.
若
x
y
z均為正數,且滿足
x+y1=5,
y+z1=2,
z+x1=1,則
x
y
z的乘積為何?
(A)1 (B)2 (C)5 (D)10
[解答]
由x+1/y=5,y+1/z=2,z+1/x=1三式相乘可得
(x+1/y)(y+1/z)(z+1/x)=5*2*1
xyz+(x+y+z)+(1/x+1/y+1/z)+1/xyz
=xyz+(x+1/y)+(y+1/z)+(z+1/x)+1/xyz
=xyz+5+2+1+1/xyz
=xyz+1/xyz+8=10
xyz+1/xyz=2
(xyz)^2-2xyz+1=0
(xyz-1)^2=0
所以xyz=1
故選(A)
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另解:
令x+1/y=5---(1),y+1/z=2---(2),z+1/x=1---(3)
由(1)得x=5-1/y=(5y-1)/y---(4)
由(2)得z=1/(2-y)---(5)
以(4)(5)代入(3)得
1/(2-y)+y/(5y-1)=1
==>(5y-1)+(2-y)y=(5y-1)(2-y)
==>-y^2+7y-1=-5y^2+11y-2
==>4y^2-4y+1=0
==>(2y-1)^2=0
==>y=1/2
代入(4)得x=(5/2-1)/(1/2)=3
代入(5)得z=1/(2-1/2)=2/3
所以xyz=3*(1/2)*(2/3)=1
故選(A)