Toán Lớp 8: Cho ab+bc+ac-a^2-b^2 -c^2=0 cmr:a=b=c đầy đủ chi tiết. ko đầy đủ, chi tiết từng bước 1=bc

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Toán Lớp 8:  Cho ab+bc+ac-a^2-b^2 -c^2=0 cmr:a=b=c đầy đủ chi tiết. ko đầy đủ, chi tiết từng bước 1=bc

hienchaudiem · Answer

$ begin{array}{l}ab+bc+ac-a^2-b^2 -c^2=0   Leftrightarrow -a^2-b^2-c^2=-ab-bc-ac   Leftrightarrow -(a^2+b^2+c^2)=-(ab+bc+ac)    Leftrightarrow a^2+b^2+c^2=ab+bc+ac    Leftrightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ac   Leftrightarrow (a^2+b^2)+(b^2+c^2)+(a^2+c^2)=2ab+2bc+2ac   Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0   Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0   Leftrightarrow  begin{cases}a-b=0  b-c=0  a-c=0 end{cases}   Leftrightarrow  begin{cases}a=b  b=c  a=c end{cases}   Leftrightarrow a=b=c  end{array}$

cobexinhdep · Answer

Theo đề b&agrave;i ta c&oacute;        ab+bc+ac-a&sup2;-b&sup2;-c&sup2;=0   &hArr;ab+bc+ca=a&sup2;+b&sup2;+c&sup2;   &hArr;a&sup2;+b&sup2;+c&sup2;=ab+bc+ca   &hArr;2a&sup2;+2.b&sup2;+2.c&sup2;=2.ab+2.bc+2.ca   &hArr;(a&sup2;+b&sup2;)+(b&sup2;+c&sup2;)+(c&sup2;+a&sup2;)=2ab+2bc+2ca   &hArr;(a&sup2;-2ab+b&sup2;)+(b&sup2;-2bc+c&sup2;)+(c&sup2;-2ca+a&sup2;)=0   &hArr;(a-b)&sup2;+(b-c)&sup2;+(c-a)&sup2;=0(*)   v&igrave; (a-b)&sup2;+(b-c)&sup2;+(c-a)&sup2;&ge;0 &forall; a,b,c   n&ecirc;n (*) xảy ra khi l&agrave; dấu '=' của bất đẳng thức(a-b)&sup2;+(b-c)&sup2;+(c-a)&sup2;&ge;0   Dấu '=' a=b=c