\(\left(a+b+c\right)^2+12=4\left(a+b+c\right)\)\(+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac+12-4\left(a+b+c\right)-2\left(ab+bc+ac\right)=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)-2\left(ab+bc+ac\right)-4\left(a+b+c\right)+12=0\)
\(\Rightarrow a^2+b^2+c^2-4a-4b-4c+12=0\)
\(\Rightarrow\left(a^2-4a+4\right)+\left(b^2-4b+4\right)+\left(c^2-4c+4\right)=0\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\)
Ta co: \(\left(a-2\right)^2\ge0\forall a\)
\(\left(b-2\right)^2\ge0\forall b\)
\(\left(c-2\right)^2\ge0\forall c\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\Leftrightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}\Leftrightarrow}a=b=c=2}\left(\right)\)
(đpcm)
Mình nghĩ thế này nhé bạn!
(a + b + c )2 + 12 = 4 (a + b +c ) + 2(ab + bc +ac)
\(\Leftrightarrow\)a2 + b2 + c2 + 2ab + 2bc + 2ac + 12 = 4a + 4b + 4c + 2ab + 2ac + 2bc
\(\Leftrightarrow\) a2 + b2 + c2 - 4a - 4b -4c +12 = 0
\(\Leftrightarrow\)a2 - 4a + 4 + b2 - 4b + 4 + c2 - 4c + 4 =0
\(\Leftrightarrow\)( a -2 )2 + (b-2)2 + (c-2)2 = 0
ta có (a-2 )2 \(\ge0\forall a\)
(b - 2 )2 \(\ge0\forall b\)
(c - 2 )2 \(\ge0\forall c\)
mà (a-2)2 + (b-2)2 + (c-2)2 = 0
\(\Rightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=2\\b=2\\c=2\end{cases}\left(đpcm\right)}\)
vậy................... khi a=b = c =2
#mã mã#
