1. Thực hiện phép tính (tính nhanh nếu có thể): a) 9/4:- 3/10; 2. Tìm x: a) 4/3:x =- 8/9;
1.
a) \[\frac{9}{4}:\frac{{ - 3}}{{10}}\] \[ = \frac{9}{4}.\frac{{10}}{{ - 3}}\] \[ = \frac{{9.10}}{{4.\left( { - 3} \right)}}\]\( = \frac{{15}}{{ - 2}}\) \( = \frac{{ - 15}}{2}\);
b) \[\frac{{ - 5}}{3}.\frac{{11}}{{25}} + \frac{{ - 5}}{3}.\frac{{14}}{{25}}\]\[ = \frac{{ - 5}}{3}\left( {\frac{{11}}{{25}} + \frac{{14}}{{25}}} \right)\]\[ = \frac{{ - 5}}{3}.\frac{{25}}{{25}} = \frac{{ - 5}}{3}\].
2.
a) \(\frac{4}{3}:x = \frac{{ - 8}}{9}\) \(x = \frac{4}{3}:\frac{{ - 8}}{9}\) \(x = \frac{4}{3}.\frac{9}{{ - 8}}\) \(x = - \frac{3}{2}\) Vậy \(x = - \frac{3}{2}\). | b) \(\frac{1}{4} - {\left( {3x - \frac{1}{2}} \right)^2} = 0\) \({\left( {3x - \frac{1}{2}} \right)^2} = \frac{1}{4}\) Trường hợp 1: \(3x - \frac{1}{2} = \frac{1}{2}\) \(3x = \frac{1}{2} + \frac{1}{2}\) \(3x = 1\) \(x = \frac{1}{3}\) Vậy \(x \in \left\{ {\frac{1}{3};0} \right\}\). | Trường hợp 2: \(3x - \frac{1}{2} = - \frac{1}{2}\) \(3x = - \frac{1}{2} + \frac{1}{2}\) \(3x = 0\) \(x = 0\) |