Ta có:
\(\left\{{}\begin{matrix}\left(x-15\right)^2=\left|x-15\right|^2\\\left|15-x\right|=\left|x-15\right|\end{matrix}\right.\)
\(\Leftrightarrow2015\left|x-15\right|+\left|x-15\right|^2-2014\left|x-15\right|=0\)
\(\Leftrightarrow\left|x-15\right|+\left|x-15\right|^2=0\Leftrightarrow\left|x-15\right|\left(\left|x-15\right|-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=16\\x=14\end{matrix}\right.\)
Đúng 0
Bình luận (1)