已知x^ y^ z^-xy-zy-xz=0 求证x=y=z
已知x^ y^ z^-xy-zy-xz=0 求证x=y=z
因为x^ y^ z^-xy-zy-xz=0
所以2(x^ y^ z^-xy-zy-xz)=0
(x-y)^2 (x-z)^2 (y-z)^2=0
而(x-y)^2,(x-z)^2,(y-z)^2均非负
所以(x-y)^2=0,
(x-z)^2=0,
(y-z)^2=0
即x=y=z
问:求证在△ABC中,已知a<(b+c)/2,求证A<(B+C)/2
答:a3/4(b^2+c^2)/(2bc)-1/4 b^2+c^2>2bc,(b^2+c^2)/2bc>1 cosA>3/4-1/4=1/2,A<60,3A<180...详情>>
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