附加题:(y-z) 2 +(x-y) 2 +(z-x) 2 =(y+z-2x) 2 +(z+x-2y) 2 +(x+y-

附加题:(y-z) 2 +(x-y) 2 +(z-x) 2 =(y+z-2x) 2 +(z+x-2y) 2 +(x+y-2z) 2
(yz+1)(zx+1)(xy+1)
(x 2 +1 )(y 2 +1 )(z 2 +1)
的值.
fangyang0916 1年前 已收到1个回答 举报

bent8221 幼苗

共回答了23个问题采纳率:95.7% 举报

∵(y-z) 2 +(x-y) 2 +(z-x) 2 =(y+z-2x) 2 +(z+x-2y) 2 +(x+y-2z) 2
∴(y-z) 2 -(y+z-2x) 2 +(x-y) 2 -(x+y-2z) 2 +(z-x) 2 -(z+x-2y) 2 =0,
∴(y-z+y+z-2x)(y-z-y-z+2x)+(x-y+x+y-2z)(x-y-x-y+2z)+(z-x+z+x-2y)(z-x-z-x+2y)=0,
∴x 2 +y 2 +z 2 -2xy-2xz-2yz=0,
∴(x-y) 2 +(x-z) 2 +(y-z) 2 =0.
∵x,y,z均为实数,
∴x=y=z.

(yz+1)(zx+1)(xy+1)
(x 2 +1 )(y 2 +1 )(z 2 +1) =
( x 2 +1)( y 2 +1)( z 2 +1)
( x 2 +1)( y 2 +1)( z 2 +1) =1.

1年前

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