Tìm x, biết: a) 0,5 - 0,6x = 0,7;   b) x + 7/3 =- 5/2

a) \(0,5 - 0,6x = 0,7\)

\(0,6x = 0,5 - 0,7\)

\(0,6x =  - 0,2\)

\(x =  - 0,2:0,6\)

\(x =  - \frac{2}{{10}}:\frac{6}{{10}}\)

\(x =  - \frac{2}{{10}}.\frac{{10}}{6}\)

\(x =  - \frac{1}{3}\)

Vậy \(x =  - \frac{1}{3}\).

b) \[x + \frac{7}{3} = \frac{{ - 5}}{2}\]

\[x = \frac{{ - 5}}{2} - \frac{7}{3}\]

\[x = \frac{{ - 29}}{6}\]

Vậy \(x = \frac{{ - 29}}{6}\).

c) \(3.{\left( {x - \frac{1}{4}} \right)^2} = \frac{{27}}{4}\)

       \({\left( {x - \frac{1}{4}} \right)^2} = \frac{{27}}{4}:3 = \frac{9}{4}\)

       \({\left( {x - \frac{1}{4}} \right)^2} = {\left( {\frac{3}{2}} \right)^2} = {\left( { - \frac{3}{2}} \right)^2}\)

Trường hợp 1:

\(x - \frac{1}{4} = \frac{3}{2}\)

\(x = \frac{3}{2} + \frac{1}{4}\)

\(x = \frac{7}{4}\)

Vậy \[x \in \left\{ {\frac{7}{4}; - \frac{5}{4}} \right\}\].

Trường hợp 2:

\(x - \frac{1}{4} =  - \frac{3}{2}\)

\(x =  - \frac{3}{2} + \frac{1}{4}\)

\(x =  - \frac{5}{4}\)