Toán 8 khó

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soccan

$ (a+b+c)^2=3(ab+ac+bc)\\
\Longleftrightarrow a^2+b^2+c^2+2(ab+bc+ac)=3(ab+ac+bc)\\
\Longleftrightarrow a^2+b^2+c^2-ab-bc-ac=0\\
\Longleftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\
\Longleftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
suy ra $a-b=b-c=c-a=0$
hay $a=b=c$
 
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