1. Thực hiện phép tính (tính nhanh nếu có thể): a) 5/7 + 4/- 14;2. Tìm x: a) 1,2x +1/2 = 0,6;
1.
a) \[\frac{5}{7} + \frac{4}{{ - 14}}\]\[ = \frac{5}{7} + \frac{{ - 2}}{7}\]\[ = \frac{3}{7}\];
b) \(\frac{2}{{11}}.\frac{{ - 5}}{4} + \frac{{ - 9}}{{11}}.\frac{5}{4} + 1\frac{3}{4}\)\( = \frac{2}{{11}}.\frac{{ - 5}}{4} + \frac{9}{{11}}.\frac{{ - 5}}{4} + \frac{7}{4}\)
\( = \frac{{ - 5}}{4}.\left( {\frac{2}{{11}} + \frac{9}{{11}}} \right) + \frac{7}{4}\)\( = \frac{{ - 5}}{4}.1 + \frac{7}{4}\)
\[ = \frac{5}{4} + \frac{7}{4}\]\[ = \frac{{12}}{4} = 3\].
2.
a) \(1,2x + \frac{1}{2} = 0,6\) \(1,2x + 0,5 = 0,6\) \(1,2x = 0,1\) \(x = \frac{1}{{12}}\). Vậy \(x = \frac{1}{{12}}\). | b) \(\left( {3x - 1} \right)\left( { - \frac{1}{2}x + 5} \right) = 0\) Suy ra \(3x - 1 = 0\) hoặc \( - \frac{1}{2}x + 5 = 0\) | |
Trường hợp 1: \(3x - 1 = 0\) \(3x = 1\) \(x = \frac{1}{3}\).
Vậy \(x \in \left\{ {\frac{1}{3};10} \right\}\). | Trường hợp 2: \( - \frac{1}{2}x + 5 = 0\) \( - \frac{1}{2}x = - 5\) \(x = - 5:\left( {\frac{{ - 1}}{2}} \right)\) \(x = 10\). | |