已知x,y,z均为实数,x>0,y>0,且a=(y-z)²/x—(z-x)²/y,b=x-y,下列结论中必定成立
问题描述：
已知x,y,z均为实数,x>0,y>0,且a=(y-z)²/x—(z-x)²/y,b=x-y,下列结论中必定成立
A.若x
答
a-b=(y-z)²/x—(z-x)²/y—(x-y)
=[y(y-z)²-x(z-x)²-xy(x-y)]/xy
∵x>0,y>0
∴分母xy>0
y(y-z)²-x(z-x)²-xy(x-y)
=y(y²-2yz+z²)-x(x²-2xz+z²)-xy(x-y)
=y³-2y²z+yz²-x³+2x²z-xz²-xy(x-y)
=(y-x)(x²+xy+y²)+2z(x²-y²)+z²(y-x)+xy(y-x)
=(y-x)(x²+xy+y²+xy)+2z(x+y)(x-y)+z²(y-x)
=(y-x)(x²+2xy+y²)-2z(x+y)(y-x)+z²(y-x)
=(y-x)[(x+y)²-2z(x+y)+z²]
=(y-x)[(x+y)²-z]²
∵x0,y>0,[(x+y)²-z]²>0
∴y-x>0
分子y(y-z)²-x(z-x)²-xy(x-y)=(y-x)[(x+y)²-z]²>0
∴a-b>0即a>b
而a=b的则是由于z,z有等于(x+y)²的情况
那么分子y(y-z)²-x(z-x)²-xy(x-y)=(y-x)[(x+y)²-z]²就为0
∴综上所述
若x