cho ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 = 4( a 2 + b 2 + c 2 -ab-bc-ca cm a=b=c

Giải thích các bước giải:

` ( a − b )^ 2 + ( b − c )^ 2 + ( c − a )^ 2 = 4( a ^2 + b^ 2 + c ^2)`

`=>a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2=4a^2+4b^2+4c^2`

`=>4a^2+4b^2+4c^2-2a^2-2b^2-2c^2-2ab-2bc-2ca=0`

`=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0`

`=>(a^2+2ab+b^2)+(b^2+2bc+c^2)+(c^2+2ca+a^2)=0`

`=> ( a − b )^ 2 + ( b − c )^ 2 + ( c − a )^ 2=0`

Có $\left\{\begin{matrix} ( a − b )^ 2\\ ( b − c )^ 2\\ ( c − a )^ 2\end{matrix}\right.$

`=> ( a − b )^ 2 + ( b − c )^ 2 + ( c − a )^ 2>=0`

Dấu `=` xảy ra `=>`$\left\{\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.$`=>a=b=c`

    Vậy `a=b=c.`