Question

cho |\(\dfrac{1}{2}\)+x|+|x-y+z|+|\(\dfrac{1}{3}\)+y|=0

tinh A=2x+y+z

help me!

Answer 1

\(\left|x+\dfrac{1}{2}\right|+\left|x-y+z\right|+\left|y+\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y+\dfrac{1}{3}=0\\x-y+z=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{1}{3}\\z=-x+y=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\end{matrix}\right.\)

\(A=2x+y+z=-1-\dfrac{1}{3}+\dfrac{1}{6}=-\dfrac{4}{3}+\dfrac{1}{6}=-\dfrac{7}{6}\)