Question

Tìm x, biết : a, |x + 2| = x b, |x - 1|- 2x = 1212 c, |0,5x - 2|-|x + 3| = 0 d, |x - 3,5|+|x + 1,3| = 0

Answer 1

Answer 2

c: Ta có: \(\left|\dfrac{1}{2}x-2\right|-\left|x+3\right|=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+3\\\dfrac{1}{2}x-2=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{-1}{2}=5\\x\cdot\dfrac{3}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-\dfrac{2}{3}\end{matrix}\right.\)