Đáp án:
B2:
d) \(x = - \dfrac{{25}}{{26}}\)
Giải thích các bước giải:
Bài 1:
\(\begin{array}{l}
a)\dfrac{3}{5}.\left( { - \dfrac{{10}}{{13}}} \right) + \dfrac{7}{{13}}\\
= - \dfrac{6}{{13}} + \dfrac{7}{{13}} = \dfrac{1}{{13}}\\
b) - \dfrac{3}{8}:{\left( { - \dfrac{1}{2}} \right)^2} + \dfrac{5}{6}.\dfrac{2}{7}\\
= - \dfrac{3}{2} + \dfrac{5}{{21}} = - \dfrac{{53}}{{42}}\\
B2:\\
a)x = - \dfrac{7}{3} + \dfrac{4}{5}\\
\to x = - \dfrac{{23}}{{15}}\\
b)x + \dfrac{2}{3} = \dfrac{1}{4} + \dfrac{5}{8}\\
\to x + \dfrac{2}{3} = \dfrac{7}{8}\\
\to x = \dfrac{5}{{24}}\\
c){\left( {3x - \dfrac{1}{2}} \right)^2} = - \dfrac{1}{{27}}\left( {vô lý} \right)\\
Do:{\left( {3x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to x \in \emptyset \\
d)\dfrac{{10x + 3x}}{{15}} = - \dfrac{5}{6}\\
\to 13x = - \dfrac{{25}}{2}\\
\to x = - \dfrac{{25}}{{26}}
\end{array}\)
