Question

cho (a-b)^2+(b-c)^2+(c-a)^2=(a+b-2c)^2+(b+c-2a)^2+(c+a-2b)^2

chứng minh a=b=c

Answer 1

(a-b)^2 + (b-c)^2 + (c-a)^2 = (a+b-2c)^2 + (b+c-2a)^2 + (c+a-2b)^2

<=> (a+b-2c)^2 - (a-b)^2 + (b+c-2a)^2 - (b-c)^2 + (c+a-2b)^2 - (c-a)^2 = 0

<=> (2b-2c)(2a-2c) + (2c-2a)(2b-2a) + (2a-2b)(2c-2b) = 0

<=> (b-c)(a-c) + (c-a)(b-a) + (a-b)(c-b) = 0

<=> ab - ac - bc + c^2 + bc - ab - ac - a^2 + ac - bc - ab + b^2 = 0

<=> a^2 + b^2 + c^2 - ab - bc - ac = 0

<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0

<=> (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) = 0

<=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0

<=> (a-b)^2=0; (b-c)^2=0; (c-a)^2=0

<=> a-b=0; b-c=0; c-a=0

<=> a=b=c (đpcm)