Giải thích các bước giải:
\(\begin{array}{l}
1.a) = 129\dfrac{5}{7}.\dfrac{5}{2} - 128\dfrac{2}{7}:\dfrac{2}{5}\\
= 129\dfrac{5}{7}.\dfrac{5}{2} - 128\dfrac{2}{7}.\dfrac{5}{2}\\
= \dfrac{5}{2}\left( {129\dfrac{5}{7} - 128\dfrac{3}{7}} \right)\\
= \dfrac{5}{2}.1\dfrac{3}{7}\\
= \dfrac{5}{2}.\dfrac{{10}}{7} = \dfrac{{25}}{7}\\
b) = \dfrac{5}{2} + \dfrac{1}{2}:\dfrac{{ - 3}}{4}.\dfrac{4}{9} - 16 - \left( { - 8} \right)\\
= \dfrac{5}{2} + \dfrac{1}{2}.\dfrac{{ - 4}}{3}.\dfrac{4}{9} - 16 + 8\\
= \dfrac{5}{2} + \dfrac{{ - 8}}{{27}} - 8 = \dfrac{{ - 313}}{{54}}\\
c) = \dfrac{{{{2.6}^9} - {2^5}{{.6}^4}{{.3}^4}}}{{{2^2}{{.6}^8}}} = \dfrac{{{{2.6}^9} - {{2.6}^4}{{.6}^4}}}{{{2^2}{{.6}^8}}}\\
= \dfrac{{{{2.6}^9} - {{2.6}^8}}}{{{2^2}{{.6}^8}}}\\
= \dfrac{{{{2.6}^8}\left( {6 - 2} \right)}}{{{2^2}{{.6}^8}}} = \dfrac{4}{2} = 2\\
2)a) - \dfrac{1}{2} + \dfrac{1}{2}x = - \dfrac{2}{3} - 3\\
- \dfrac{1}{2} + \dfrac{1}{2}x = - \dfrac{{11}}{3}\\
\dfrac{1}{2}x = - \dfrac{{11}}{3} + \dfrac{1}{2}\\
\dfrac{1}{2}x = - \dfrac{{19}}{6}\\
x = - \dfrac{{19}}{3}\\
b)TH1:x + \dfrac{1}{2} = \dfrac{1}{3}\\
x = \dfrac{1}{3} - \dfrac{1}{2}\\
x = - \dfrac{1}{6}\\
TH2:x + \dfrac{1}{2} = - \dfrac{1}{3}\\
x = - \dfrac{1}{3} - \dfrac{1}{2}\\
x = - \dfrac{5}{6}
\end{array}\)
