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

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

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

\(x = \frac{3}{2} + \frac{3}{4} = \frac{6}{4} + \frac{3}{4}\)

\(x = \frac{9}{4}\)

Vậy \(x = \frac{9}{4}\).

b) \(\frac{4}{7} - \frac{3}{7} \cdot \left( {x + \frac{1}{9}} \right) = \frac{1}{7}\)

\(\frac{3}{7} \cdot \left( {x + \frac{1}{9}} \right) = \frac{4}{7} - \frac{1}{7}\)

\(\frac{3}{7} \cdot \left( {x + \frac{1}{9}} \right) = \frac{3}{7}\)

\(x + \frac{1}{9} = 1\)

\(x = 1 - \frac{1}{9}\)

\(x = \frac{8}{9}\)

Vậy \(x = \frac{8}{9}\).

c) \(\frac{3}{2}{\left( {x + \frac{1}{2}} \right)^2} - 1,5:0,9 = 4\frac{1}{3}\)

\(\frac{3}{2} \cdot {\left( {x + \frac{1}{2}} \right)^2} - \frac{3}{2}:\frac{9}{{10}} = \frac{{13}}{3}\)

\(\frac{3}{2} \cdot {\left( {x + \frac{1}{2}} \right)^2} - \frac{3}{2} \cdot \frac{{10}}{9} = \frac{{13}}{3}\)

\(\frac{3}{2} \cdot {\left( {x + \frac{1}{2}} \right)^2} - \frac{5}{3} = \frac{{13}}{3}\)

\(\frac{3}{2} \cdot {\left( {x + \frac{1}{2}} \right)^2} = \frac{{13}}{3} + \frac{5}{3}\)

\(\frac{3}{2} \cdot {\left( {x + \frac{1}{2}} \right)^2} = 6\)

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

TH1: \(x + \frac{1}{2} = 2\)

\(x = 1,5\)

TH 2: \(x + \frac{1}{2} = - 2\)

\(x = - 2,5\)

Vậy \(x \in \left\{ {1,5;\,\,2,5} \right\}.\)

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