Đáp án:
B2:
\(x = \dfrac{{122}}{{495}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
B = \left\{ {\left[ {\dfrac{5}{2} - 4\left( {\dfrac{5}{2} - \dfrac{6}{5}} \right) + \dfrac{3}{8}} \right]:\left[ {4\left( {\dfrac{5}{2} - \dfrac{6}{5}} \right) - \dfrac{3}{5}:\dfrac{2}{5}} \right]} \right\} - \dfrac{{55}}{{148}}\\
= \dfrac{{\dfrac{5}{2} - 4.\dfrac{{13}}{{10}} + \dfrac{3}{8}}}{{4.\dfrac{{13}}{{10}} - \dfrac{3}{2}}} - \dfrac{{55}}{{148}}\\
= \dfrac{{ - \dfrac{{26}}{5} + \dfrac{{23}}{8}}}{{\dfrac{{26}}{5} - \dfrac{3}{2}}} - \dfrac{{55}}{{148}}\\
= - \dfrac{{93}}{{40}}:\dfrac{{37}}{{10}} - \dfrac{{55}}{{148}}\\
= - \dfrac{{93}}{{148}} - \dfrac{{55}}{{148}} = - \dfrac{{148}}{{148}} = - 1\\
B2:\\
\left( {\dfrac{{12}}{{25}} + \left( {\dfrac{3}{4} + \dfrac{7}{4}} \right) - 1} \right)x = \dfrac{7}{5} - \dfrac{3}{{25}}.\left( {\dfrac{{53}}{5} - \dfrac{3}{5}} \right)\\
\to \left( {\dfrac{{12}}{{25}} + \dfrac{5}{2} - 1} \right)x = \dfrac{7}{5} - \dfrac{3}{{25}}.\dfrac{{38}}{5}\\
\to \dfrac{{99}}{{50}}.x = \dfrac{7}{5} - \dfrac{{114}}{{125}}\\
\to \dfrac{{99}}{{50}}.x = \dfrac{{61}}{{125}}\\
\to x = \dfrac{{122}}{{495}}
\end{array}\)
