Tìm \(x\), biết:
a) \(x + \frac{{11}}{{12}} = \frac{{23}}{{24}}\)
b) \(\frac{{11}}{8} - \frac{3}{8} \cdot x = \frac{1}{8}\)
c) \({\left( {{\rm{x}} - \frac{1}{2}} \right)^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{4}\)
Dựa vào quy tắc tính với phân số.
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) \({\left( {{\rm{x}} - \frac{1}{2}} \right)^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{4}\)
\(\begin{array}{l}\left[ \begin{array}{l}{\rm{x}} - \frac{1}{2}{\rm{ = }}\frac{{\rm{1}}}{2}\\{\rm{x}} - \frac{1}{2}{\rm{ = }}\frac{{ - 1}}{2}\end{array} \right.\\\left[ \begin{array}{l}x = \frac{{\rm{1}}}{2} + \frac{1}{2}\\x = \frac{{ - 1}}{2} + \frac{1}{2}\end{array} \right.\\\left[ \begin{array}{l}x = 1\\x = 0\end{array} \right.\end{array}\)
Vậy \(x = 1;x = 0\).