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
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![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](/uploads/image/z/5222212-52-2.jpg?t=%E8%AE%BEa%E7%9A%84x%E6%AC%A1%E6%96%B9%3Db%E7%9A%84y%E6%AC%A1%E6%96%B9%3D%EF%BC%88ab%EF%BC%89%E7%9A%84z%E6%AC%A1%E6%96%B9%2C%E4%B8%94xyz%E4%B8%8D%E7%AD%89%E4%BA%8E0%2Ca%E5%92%8Cb%E5%9D%87%E4%B8%BA%E4%B8%8D%E7%AD%89%E4%BA%8E1%E7%9A%84%E6%AD%A3%E6%95%B0%2C%E8%AF%81%E6%98%8Ez%3Dx%2By%E5%88%86%E4%B9%8Bxy)
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
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
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
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
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)
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
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)