Tìm \(x\), biết:

a) \(x + \frac{{11}}{{12}} = \frac{{23}}{{24}}\)

b) \(\frac{{11}}{8} - \frac{3}{8} \cdot x = \frac{1}{8}\)

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

a) \(x + \frac{{11}}{{12}} = \frac{{23}}{{24}}\)

\(\begin{array}{l}x = \frac{{23}}{{24}} - \frac{{11}}{{12}}\\x = \frac{{23}}{{24}} - \frac{{22}}{{24}}\\x = \frac{1}{{24}}\end{array}\)

Vậy \(x = \frac{1}{{24}}\)

b) \(\frac{{11}}{8} - \frac{3}{8} \cdot x = \frac{1}{8}\)

\(\begin{array}{l}\frac{{11}}{8} - \frac{3}{8} \cdot x = \frac{1}{8}\\\frac{3}{8}x = \frac{{11}}{8} - \frac{1}{8}\\\frac{3}{8}x = \frac{5}{4}\\x = \frac{5}{4}:\frac{3}{8}\\x = \frac{{10}}{3}\end{array}\)

Vậy \(x = \frac{{10}}{3}\)

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

                  \(\begin{array}{l}x = \frac{1}{2} + \frac{1}{2}\\x = 1\end{array}\)

TH2: \(x - \frac{1}{2} =  - \frac{1}{2}\)

                  \(\begin{array}{l}x =  - \frac{1}{2} + \frac{1}{2}\\x = 0\end{array}\)

Vậy \(x = 1;x = 0\).