1. Thực hiện phép tính (tính nhanh nếu có thể): a) (1/8- 9/8.4/3); 2. Tìm x: a)3/7 + x = 4/5;
1.
a) \(\frac{1}{8} - \frac{9}{8}.\frac{4}{3}\)\( = \frac{1}{8} - \frac{3}{2}\)\( = \frac{1}{8} - \frac{{12}}{8}\)\( = \frac{{ - 11}}{8}\);
b) \[\frac{{ - 3}}{4}.\frac{{2022}}{{2023}} - \frac{{2022}}{{2023}}.\frac{1}{4}\]\[ = \frac{{2022}}{{2023}}.\left( {\frac{{ - 3}}{4} - \frac{1}{4}} \right)\]
\[ = \frac{{2022}}{{2023}}.\left( { - 1} \right)\]\[ = - \frac{{2022}}{{2023}}\].
2.
a) \[\frac{3}{7} + x = \frac{4}{5}\] \[x = \frac{4}{5} - \frac{3}{7}\] \[x = \frac{{28}}{{35}} - \frac{{15}}{{35}}\] \[x = \frac{{13}}{{35}}\] Vậy \[x = \frac{{13}}{{35}}\]. | b) \[\left( {3x - 1} \right)\left( {\frac{{ - 1}}{2}x + 5} \right) = 0\] • TH1: \[3x--1 = 0\] \[3x = 1\] \[x = \frac{1}{3}\]. • TH2: \[\frac{{ - 1}}{2}x + 5 = 0\] \[\frac{{ - 1}}{2}x = - 5\] \[x = - 5:\left( {\frac{{ - 1}}{2}} \right)\] \[x = 10\]. Vậy \[x \in \left\{ {\frac{1}{3};\,\,10} \right\}\]. |