Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left| {x + \dfrac{1}{2}} \right| \ge 0,\,\,\,\forall x\\
\left| {3 - y} \right| \ge 0,\,\,\forall y\\
\Rightarrow \left| {x + \dfrac{1}{2}} \right| + \left| {3 - y} \right| \ge 0,\,\,\forall x,y\\
\left| {x + \dfrac{1}{2}} \right| + \left| {3 - y} \right| = 0 \Leftrightarrow \left\{ \begin{array}{l}
\left| {x + \dfrac{1}{2}} \right| = 0\\
\left| {3 - y} \right| = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{1}{2} = 0\\
3 - y = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{1}{2}\\
y = 3
\end{array} \right.\\
b,\\
\left| {\dfrac{1}{2} - \dfrac{1}{3} + x} \right| = - \dfrac{1}{4} - \left| y \right|\\
\Leftrightarrow \left| {\dfrac{1}{6} + x} \right| + \left| y \right| = - \dfrac{1}{4}\\
\left| {\dfrac{1}{6} + x} \right| \ge 0,\,\,\forall x\\
\left| y \right| \ge 0,\,\,\forall y\\
\Rightarrow \left| {\dfrac{1}{6} + x} \right| + \left| y \right| \ge 0,\,\,\,\forall x,y\\
\Rightarrow ptvn\\
c,\\
\left| {x - y} \right| \ge 0,\,\,\,\forall x,y\\
\left| {y + \dfrac{9}{{25}}} \right| \ge 0,\,\,\forall y\\
\Rightarrow \left| {x - y} \right| + \left| {y + \dfrac{9}{{25}}} \right| \ge 0,\,\,\,\forall x,y\\
\left| {x - y} \right| + \left| {y + \dfrac{9}{{25}}} \right| = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\left| {x - y} \right| = 0\\
\left| {y + \dfrac{9}{{25}}} \right| = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
y + \dfrac{9}{{25}} = 0
\end{array} \right. \Leftrightarrow x = y = - \dfrac{9}{{25}}
\end{array}\)