Đề bài

Tìm x, biết: a)\( - \frac{3}{5}.x = \frac{{12}}{{25}};\) b)\(\frac{3}{5}x - \frac{3}{4} =  - 1\frac{1}{2};\) c)\(\frac{2}{5} + \frac{3}{5}:x = 0,5;\) d)\(\frac{3}{4} - \left( {x - \frac{1}{2}} \right) = 1\frac{2}{3}\) e)\(2\frac{2}{{15}}:\left( {\frac{1}{3} - 5x} \right) =  - 2\frac{2}{5}\) g)\({x^2} + \frac{1}{9} = \frac{5}{3}:3.\)

Lời giải chi tiết

a)\(\begin{array}{l} - \frac{3}{5}.x = \frac{{12}}{{25}}\\x = \frac{{12}}{{25}}:\frac{{ - 3}}{5}\\x = \frac{{12}}{{25}}.\frac{{ - 5}}{3}\\x = \frac{{ - 4}}{5}\end{array}\)Vậy \(x = \frac{{ - 4}}{5}\)b)\(\begin{array}{l}\frac{3}{5}x - \frac{3}{4} =  - 1\frac{1}{2};\\\frac{3}{5}x = \frac{{ - 3}}{2} + \frac{3}{4}\\\frac{3}{5}x = \frac{{ - 3}}{4}\\x = \frac{{ - 3}}{4}:\frac{3}{5}\\x = \frac{{ - 3}}{4}.\frac{5}{3}\\x = \frac{{ - 5}}{4}\end{array}\)Vậy \(x = \frac{{ - 5}}{4}\).c)\(\begin{array}{l}\frac{2}{5} + \frac{3}{5}:x = 0,5\\\frac{3}{5}:x = \frac{1}{2} - \frac{2}{5}\\\frac{3}{5}:x = \frac{1}{{10}}\\x = \frac{3}{5}:\frac{1}{{10}}\\x = \frac{3}{5}.10\\x = 6\end{array}\)Vậy \(x = 6\).d)\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{1}{2}} \right) = 1\frac{2}{3}\\x - \frac{1}{2} = \frac{3}{4} - \frac{5}{3}\\x - \frac{1}{2} = \frac{{ - 11}}{{12}}\\x = \frac{{ - 11}}{{12}} + \frac{1}{2}\\x = \frac{{ - 5}}{{12}}\end{array}\)Vậy \(x = \frac{{ - 5}}{{12}}\).e)\(\begin{array}{l}2\frac{2}{{15}}:\left( {\frac{1}{3} - 5x} \right) =  - 2\frac{2}{5}\\\frac{{32}}{{15}}:\left( {\frac{1}{3} - 5x} \right) =  - \frac{{12}}{5}\\\frac{1}{3} - 5x = \frac{{32}}{{15}}:\frac{{ - 12}}{5}\\\frac{1}{3} - 5x = \frac{{32}}{{15}}.\frac{{ - 5}}{12}\\\frac{1}{3} - 5x = \frac{{ - 8}}{9}\\5x = \frac{1}{3} + \frac{8}{9}\\5x = \frac{{11}}{9}\\x = \frac{{11}}{9}:5\\x = \frac{{11}}{{45}}\end{array}\) Vậy \(x = \frac{{11}}{{45}}\).g)\({x^2} + \frac{1}{9} = \frac{5}{3}:3\\{x^2} + \frac{1}{9} = \frac{5}{9}\\{x^2} = \frac{5}{9} - \frac{1}{9}\\{x^2} = \frac{4}{9}\\{x^2} = (\frac{2}{3})^2\\x = \frac{2}{3}\,\ hoặc \,\ x = \frac{-2}{3}\)Vậy \(x \in \{\frac{2}{3};\frac{-2}{3}\}\).