Tìm \(x\), biết:

    a) \(5,4 - x =  - 1,2\);                                   b) \(\frac{4}{3}:x = \frac{{ - 8}}{9}\);

    c) \(1\frac{3}{4} - 2.\left( {\frac{1}{2} + x} \right) = \frac{3}{2}\); d) \(\frac{1}{4} - {\left( {3x - \frac{1}{2}} \right)^2} = 0\).

a) \(5,4 - x =  - 1,2\)

\(x = 5,4 - \left( { - 1,2} \right)\)

\(x = 5,4 + 1,2\)

\(x = 6,6\)

Vậy \(x = 6,6\).

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

\(2\left( {\frac{1}{2} + x} \right) = \frac{7}{4} - \frac{3}{2}\)

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

\(\frac{1}{2} + x = \frac{1}{8}\)

\(x = \frac{1}{8} - \frac{1}{2}\)

\(x = \frac{{ - 3}}{8}\)

Vậy \(x = \frac{{ - 3}}{8}\).

b) \(\frac{4}{3}:x = \frac{{ - 8}}{9}\)

\(x = \frac{4}{3}:\frac{{ - 8}}{9}\)

\(x = \frac{4}{3}.\frac{9}{{ - 8}}\)

\(x =  - \frac{3}{2}\)

Vậy \(x =  - \frac{3}{2}\).

d) \(\frac{1}{4} - {\left( {3x - \frac{1}{2}} \right)^2} = 0\)

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

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 1:

\(3x - \frac{1}{2} =  - \frac{1}{2}\)

\(3x =  - \frac{1}{2} + \frac{1}{2}\)

\(3x = 0\)

\(x = 0\)