Đáp án:
c) \(\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne 1\\
\dfrac{{x - 1}}{4} = \dfrac{9}{{x - 1}}\\
\to {\left( {x - 1} \right)^2} = 36\\
\to \left| {x - 1} \right| = 6\\
\to \left[ \begin{array}{l}
x - 1 = 6\\
x - 1 = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = - 5
\end{array} \right.\\
b) - \dfrac{1}{2} - \dfrac{{x - 1}}{3} = \dfrac{3}{4}\\
\to \dfrac{{ - 6 - 4x + 4 - 9}}{{12}} = 0\\
\to - 4x - 11 = 0\\
\to x = - \dfrac{{11}}{4}\\
c)\dfrac{1}{3} - {\left( {x - \dfrac{1}{2}} \right)^2} = - \dfrac{2}{3}\\
\to 1 = {\left( {x - \dfrac{1}{2}} \right)^2}\\
\to \left| {x - \dfrac{1}{2}} \right| = 1\\
\to \left[ \begin{array}{l}
x - \dfrac{1}{2} = 1\\
x - \dfrac{1}{2} = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
d)2x - \dfrac{{11}}{3} = \dfrac{x}{2} + \dfrac{5}{4}\\
\to \dfrac{3}{2}x = \dfrac{{16}}{3}\\
\to x = \dfrac{{32}}{9}\\
e)3x - \dfrac{3}{2} + 2x + \dfrac{1}{2} = - \dfrac{3}{2}\\
\to 5x = - \dfrac{1}{2}\\
\to x = - \dfrac{1}{{10}}\\
f)\dfrac{1}{2}x - \dfrac{3}{2} - 2x + \dfrac{2}{3} = - \dfrac{5}{2}\\
\to - \dfrac{3}{2}x = - \dfrac{5}{3}\\
\to x = \dfrac{{10}}{9}\\
g)\dfrac{1}{2} - 2x + 2 = - \dfrac{1}{3}\\
\to - 2x = - \dfrac{{17}}{6}\\
\to x = \dfrac{{17}}{{12}}
\end{array}\)
