Ta có: \(a^2+b^2+c^2+d^2=a\left(b+c+d\right)\)
\(\Rightarrow a^2+b^2+c^2+d^2-ab-ac-ad=0\)
\(\Rightarrow4a^2+4b^2+4c^2+4d^2-4ab-4ac-4ad=0\)
\(\Rightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a-4ad+4d^2\right)+a^2=0\)
\(\Rightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+a^2=0\)
Vì \(\left(a-2b\right)^2\ge0;\left(a-2c\right)^2\ge0;\left(a-2d\right)^2\ge0;a^2\ge0\)
\(\Rightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+a^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=0\\\left(a-2b\right)^2=0\\\left(a-2c\right)^2=0\\\left(a-2d\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\0-2b=0\\0-2c=0\\0-2d=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\2b=0\\2c=0\\2d=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\b=0\\c=0\\d=0\end{matrix}\right.\)
Vậy a=b=c=0
