Đáp án:
$\begin{array}{l}
1)b)4.{\left( {\dfrac{{ - 1}}{2}} \right)^3} - 2.{\left( {\dfrac{{ - 1}}{2}} \right)^2} + 3.\left( { - \dfrac{1}{2}} \right) + 1\\
= 4.\left( {\dfrac{{ - 1}}{8}} \right) - 2.\dfrac{1}{4} - \dfrac{3}{2} + 1\\
= \dfrac{{ - 1}}{2} - \dfrac{1}{2} - \dfrac{3}{2} + 1\\
= - 1 - \dfrac{3}{2} + 1\\
= - \dfrac{3}{2}\\
B4)\\
\dfrac{a}{b} = \dfrac{c}{d} = k\left( {k \ne 0} \right)\\
\Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
\Rightarrow \dfrac{{2a - 7b}}{{4a}} = \dfrac{{2.b.k - 7b}}{{4.b.k}} = \dfrac{{2k - 7}}{{4k}}\\
\dfrac{{2c - 7d}}{{4c}} = \dfrac{{2.d.k - 7d}}{{4.d.k}} = \dfrac{{2k - 7}}{{4k}}\\
\Rightarrow \dfrac{{2a - 7b}}{{4a}} = \dfrac{{2c - 7d}}{{4c}}\left( { = \dfrac{{2k - 7}}{{4k}}} \right)
\end{array}$
