Đáp án:
B2:
c) \(\left[ \begin{array}{l}
x = - \dfrac{{23}}{{35}}\\
x = - \dfrac{{33}}{{35}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)\dfrac{1}{3}.\dfrac{{131}}{5} - \dfrac{3}{4}.\dfrac{{221}}{5}\\
= \dfrac{{131}}{{15}} - \dfrac{{663}}{{20}} = - \dfrac{{293}}{{12}}\\
b)\dfrac{3}{5} - \dfrac{2}{3} + \dfrac{5}{2} = - \dfrac{1}{{15}} + \dfrac{5}{2}\\
= \dfrac{{73}}{{30}}\\
c)\dfrac{{15}}{2}.\left( { - \dfrac{3}{5}} \right) + \dfrac{5}{2}.\left( { - \dfrac{3}{5}} \right)\\
= \left( { - \dfrac{3}{5}} \right).\left( {\dfrac{{15 + 5}}{2}} \right)\\
= \left( { - \dfrac{3}{5}} \right).10 = - 6\\
B2:\\
a)\dfrac{3}{2}x = - 6\\
\to 3x = - 12\\
\to x = - 4\\
b)\dfrac{2}{5}x = - \dfrac{4}{{15}}\\
\to x = - \dfrac{2}{3}\\
c)\left| {x + \dfrac{4}{5}} \right| = \dfrac{1}{7}\\
\to \left[ \begin{array}{l}
x + \dfrac{4}{5} = \dfrac{1}{7}\\
x + \dfrac{4}{5} = - \dfrac{1}{7}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{{23}}{{35}}\\
x = - \dfrac{{33}}{{35}}
\end{array} \right.
\end{array}\)
