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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: CMR nếu: $(x-y)^{2}$ + $(y-z)^{2}$ + $(z-x)^{2}$ $=$ $(y+z-2x)^{2}$ + $(z+x+2y)^{2}$ + $(x+y-2z)^{2}$ thì $x=y=z$

Toán Lớp 8: CMR nếu: $(x-y)^{2}$ + $(y-z)^{2}$ + $(z-x)^{2}$ $=$ $(y+z-2x)^{2}$ + $(z+x+2y)^{2}$ + $(x+y-2z)^{2}$ thì $x=y=z$

Comments ( 2 )

  1. $\\$
    giả sử x=y=z
    (x-y)^2+(y-z)^2+(z-x)^2
    =(y-y)^2+(y-y)^2+(x-x)^2
    = 0 (1)
    (y+z-2x)^2+(z+x-2y)^2 + (x+y-2z)^2
    =(x+x-2x)^2+(x+x-2x)^2 + (x+x-2x)^2
    = 0 (2)
    (1)(2)
    => (x-y)^2+(y-z)^2+(z-x)^2=(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2
    Vậy nếu (x-y)^2+(y-z)^2+(z-x)^2=(y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2 thì x=y=z
     

  2. Sửa đề : (x-y)^2 + (y+z)^2 + (z-x)^2 = (y+z-2x)^2 + (z+x-2y)^2 + (x+y-2z)^2 thì x = y =z
    Giả xử x =y =z
    Từ đó ta được :
    $\\$
    (x-y)^2 +(y-z)^2 + (z-x)^2 = (x-x)^2 + (y-y)^2 + (z-z)^^2 =0
    $\\$
    (y+z-2x)^2 + (z+x+2y)^2 + (x+y-2z)^2 = ( y+y -2y)^2 + (z+z-2z)^2 + (x+x- 2x)^2 = 0
    $\\$
    Nhận thấy 2 vế giả thiết cho đều bằng nhau ( =0) .
    Vậy (x-y)^2 + (y+z)^2 + (z-x)^2 = (y+z-2x)^2 + (z+x-2y)^2 + (x+y-2z)^2 <=> x = y =z

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222-9+11+12:2*14+14 = ? ( )