Toán Lớp 8: cho a^2+b^2+c^2=ab+bc+ac.CMR:a=b=c

Question

Toán Lớp 8:  cho a^2+b^2+c^2=ab+bc+ac.CMR:a=b=c

thaolanphuobg · Answer

$ text{Giải đáp:}$     $a^2+b^2+c^2=ab+bc+ac$     $&hArr;2a^2+2b^2+2c^2=2ab+2bc+2ac$     $&hArr;(a^2-2ab+b^2)+(b^2-2bc+c^2 )+( c^2-2ac+a^2)=0$     $&hArr;(a-b)^2+(b-c)^2+(c-a)^2=0$     Ta thấy :     $(a-b)^2&ge;0&forall;a;b$     $(b-c)^2&ge;0&forall;c;b$     $(c-a)^2&ge;0&forall;c;a$     $&rArr;(a-b)^2+(b-c)^2+(c-a)^2&ge;0&forall;a;b;c$     Dấu '=' xảy ra &hArr;a=b=c (đpcm)

ngocbichchau · Answer

a^2 + b^2 + c^2 = ab + bc + ac   &hArr; a^2 + b^2 + c^2 - ab - bc -  ac  = 0   &hArr; 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc -  2ac = 0   &hArr; a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - ca + a^2 = 0   &hArr; (a - b)^2 + (b - a)^2 + (c - a)^2 = 0    Ta c&oacute; :   (a - b)^2 + (b - a)^2 + (c - a)^2 &ge; 0   m&agrave; (a - b)^2 + (b - a)^2 + (c - a)^2 = 0    &rArr; {((a - b)^2=0),((b - a)^2=0),((c - a)^2=0):}   &hArr;{(a - b=0),(b - a=0),(c - a=0):}   &hArr; {(a = b),(b = a),(c = a):}   &hArr; a=b=c