a) \(\left|x-2017\right|+\left|y+2018\right|=0\)
ta có : \(\left|x-2017\right|\ge0\) với mọi \(x\) và \(\left|y+2018\right|\ge0\) với mọi \(y\)
\(\Rightarrow\left|x-2017\right|+\left|y+2018\right|\ge0\) với mọi \(x;y\)
\(\Rightarrow\) \(\left|x-2017\right|+\left|y+2018\right|=0\) \(\Leftrightarrow\left[{}\begin{matrix}\left|x-2017\right|=0\\\left|y+2018\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2017=0\\y+2018=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2017\\y=-2018\end{matrix}\right.\) vậy \(x=2017;y=-2018\)
b) \(\left|2x+3\right|=6-x\)
th1: \(2x+3\ge0\Leftrightarrow2x\ge-3\Leftrightarrow x\ge\dfrac{-3}{2}\)
\(\Rightarrow\left|2x+3\right|=6-x\Leftrightarrow2x+3=6-x\Leftrightarrow2x+x=6-3\)
\(\Leftrightarrow3x=3\Leftrightarrow x=1\left(tmđk\right)\)
th2: \(2x+3< 0\Leftrightarrow2x< -3\Leftrightarrow x< \dfrac{-3}{2}\)
\(\Rightarrow\left|2x+3\right|=6-x\Leftrightarrow-\left(2x+3\right)=6-x\Leftrightarrow-2x-3=6-x\)
\(\Leftrightarrow-2x+x=6+3\Leftrightarrow-x=9\Leftrightarrow x=-9\left(tmđk\right)\)
vậy \(x=1;x=-9\)
