设a的x次方=b的y次方=（ab）的z次方，且xyz不等于0，a和b均为不等于1的正数，证明z=x+y分之xy

Question

如题所述

Answer 1

a^x=(ab)^z=a^z*b^z
a^(x-z)=b^z
b=a^[(x-z)/z] (1)
b^y=(ab)^z=a^z*b^z
b^(y-z)=a^z
b=a^[z/(y-z)] (2)
(1)=(2)
所以a^[(x-z)/z]=a^[z/(y-z)]
即(x-z)/z=z/(y-z)
z²=(x-z)(y-z)=xy-xz-yz+z²
z(x+y)=xy
故z=xy/(x+y)
得证

Answer 2

由题得，a^x = b^y = a^z × b^z
a^x = a^z × b^z
可得，a^x ÷ a^z = b^z
a^(x - z) = b^z
又∵a^x = b^y
∴ (x - z) ：x = z ：y
zx = xy - zy
z(x + y) = xy
z = xy ÷ (x + y)

Answer 3

a^x=(ab)^z=a^z*b^z
a^(x-z)=b^z
b=a^[(x-z)/z] (1)
b^y=(ab)^z=a^z*b^z
b^(y-z)=a^z
b=a^[z/(y-z)] (2)
(1)=(2)
所以a^[(x-z)/z]=a^[z/(y-z)]
即(x-z)/z=z/(y-z)
z²=(x-z)(y-z)=xy-xz-yz+z²
z(x+y)=xy
故z=xy/(x+y)

Answer 4

你这道题因该是高一的吧，对数函数中的。
假设a^x=b^y=（ab）^z=k,所以x=loga(k),y=logb(k),z=logab(k),所以1/z=logk(ab)=logk(a)+logk(b)=1/x+1/y=(x+y)/xy,所以z=x+y分之xy