Exercise 2

The expresion:

\[ \begin{aligned} &\frac{1}{2} - \left[ \frac{1}{2} - \left\{ \frac{1}{4} - \left( 2 \cdot \left( \frac{1}{2} \right) - \frac{1}{4} \right) \right\} \right. \\[.5em] &\hspace{9em} \left. + \frac{1}{2} - \left( - \frac{1}{4} \right) \right] \end{aligned} \]
\[ \frac{1}{2} - \left[ \frac{1}{2} - \left\{ \frac{1}{4} - \left( 2 \cdot \left( \frac{1}{2} \right) - \frac{1}{4} \right) \right\} + \frac{1}{2} - \left( - \frac{1}{4} \right) \right] \]

is equivalent to:

The expresion:   \( \cfrac{1}{2} - \left[ \cfrac{1}{2} - \left\{ \cfrac{1}{4} - \left( 2 \cdot \left( \cfrac{1}{2} \right) - \cfrac{1}{4} \right) \right\} + \cfrac{1}{2} - \left( - \cfrac{1}{4} \right) \right] \),   is equivalent to:

  1. \(\cfrac{3}{4}\)

  2. \(-\cfrac{5}{4}\)

  3. \(-\cfrac{7}{4}\)

  4. \(-\cfrac{1}{4}\)

  5. \(\cfrac{3}{2}\)

Try to solve it before checking the answer.

  1. \(-\cfrac{5}{4}\)

By eliminating the grouping symbols, from outside to inside, we obtain:

\[
\begin{aligned}
&\frac{1}{2} – \left[ \frac{1}{2} – \left\{ \frac{1}{4} – \left( 2 \cdot \left( \frac{1}{2} \right) – \frac{1}{4} \right) \right\}
\right.
\\[.5em]
&\hspace{1em}
\left.
+ \frac{1}{2} – \left( – \frac{1}{4} \right)
\right] =
\\[.5em]
&\hspace{1em}
\frac{1}{2} – \frac{1}{2} + \left\{
\frac{1}{4} – \left( 2 \cdot \left(
\frac{1}{2}
\right) – \frac{1}{4}
\right)
\right\}
\\[.5em]
&\hspace{1em}- \frac{1}{2} + \left( – \frac{1}{4} \right) =
\\[.5em]
&\hspace{1em}
\frac{1}{2} – \frac{1}{2} + \frac{1}{4} – 2 \cdot \left( \frac{1}{2} \right) + \frac{1}{4}
\\[.5em]
&\hspace{3em} – \frac{1}{2} – \frac{1}{4} =
\\[.5em]
&
\frac{1}{2} – \frac{1}{2} + \frac{1}{4} – \frac{2}{2} + \frac{1}{4} – \frac{1}{2} – \frac{1}{4}
\\[.5em]
&\hspace{7em}
=
-\frac{3}{2} + \frac{1}{4}
\\[.5em]
&\hspace{7em}
= \frac{ -6 + 1 }{4}
\\[.5em]
&\hspace{7em}
= \frac{-5}{4}
\\[.5em]
&\hspace{7em}
= \boldsymbol{ -\frac{5}{4} }
\end{aligned}
\]
\[
\begin{aligned}
&\frac{1}{2} – \left[ \frac{1}{2} – \left\{ \frac{1}{4} – \left( 2 \cdot \left( \frac{1}{2} \right) – \frac{1}{4} \right) \right\}
+ \frac{1}{2} – \left( – \frac{1}{4} \right)
\right]
\\[.5em]
&\hspace{1em}
=
\frac{1}{2} – \frac{1}{2} + \left\{
\frac{1}{4} – \left( 2 \cdot \left(
\frac{1}{2}
\right) – \frac{1}{4}
\right)
\right\}
– \frac{1}{2} + \left( – \frac{1}{4} \right)
\\[.5em]
&\hspace{1em}
=
\frac{1}{2} – \frac{1}{2} + \frac{1}{4} – 2 \cdot \left( \frac{1}{2} \right) + \frac{1}{4}
– \frac{1}{2} – \frac{1}{4}
\\[.5em]
&\hspace{1em}
=
\frac{1}{2} – \frac{1}{2} + \frac{1}{4} – \frac{2}{2} + \frac{1}{4} – \frac{1}{2} – \frac{1}{4}
\\[.5em]
&\hspace{1em}
=
-\frac{3}{2} + \frac{1}{4}
\\[.5em]
&\hspace{1em}
= \frac{ -6 + 1 }{4}
\\[.5em]
&\hspace{1em}
= \frac{-5}{4}
\\[.5em]
&\hspace{1em}
= \boldsymbol{ -\frac{5}{4} }
\end{aligned}
\]
\[
\begin{aligned}
\frac{1}{2} – \left[ \frac{1}{2} – \left\{ \frac{1}{4} – \left( 2 \cdot \left( \frac{1}{2} \right) – \frac{1}{4} \right) \right\}
+ \frac{1}{2} – \left( – \frac{1}{4} \right)
\right]
&
=
\frac{1}{2} – \frac{1}{2} + \left\{
\frac{1}{4} – \left( 2 \cdot \left(
\frac{1}{2}
\right) – \frac{1}{4}
\right)
\right\}
– \frac{1}{2} + \left( – \frac{1}{4} \right)
\\[.5em]
&
=
\frac{1}{2} – \frac{1}{2} + \frac{1}{4} – 2 \cdot \left( \frac{1}{2} \right) + \frac{1}{4}
– \frac{1}{2} – \frac{1}{4}
\\[.5em]
&
=
\frac{1}{2} – \frac{1}{2} + \frac{1}{4} – \frac{2}{2} + \frac{1}{4} – \frac{1}{2} – \frac{1}{4}
\\[.5em]
&
=
-\frac{3}{2} + \frac{1}{4}
\\[.5em]
&
= \frac{ -6 + 1 }{4}
\\[.5em]
&
= \frac{-5}{4}
\\[.5em]
&
= \boldsymbol{ -\frac{5}{4} }
\end{aligned}
\]

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