已知a/x=b/y=c/z,证a²/x²=b²/y²=c²/z²=(a+b+c)²/(x+y+z)²分式的恒等变形
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已知a/x=b/y=c/z,证a²/x²=b²/y²=c²/z²=(a+b+c)²/(x+y+z)²分式的恒等变形
已知a/x=b/y=c/z,证a²/x²=b²/y²=c²/z²=(a+b+c)²/(x+y+z)²
分式的恒等变形
已知a/x=b/y=c/z,证a²/x²=b²/y²=c²/z²=(a+b+c)²/(x+y+z)²分式的恒等变形
证明如下:
设a/x=b/y=c/z=k
则a=kx,b=ky,c=kz
所以
a²/x²=(kx)²/x²=k²
b²/y²=(ky)²/y²=k²
c²/z²=(kz)²/z²=k²
(a+b+c)²/(x+y+z)²=(kx+ky+kz)²/(x+y+z)²=k²(x+y+z)²/(x+y+z)²=k²
故a²/x²=b²/y²=c²/z²=(a+b+c)²/(x+y+z)²
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