Đáp án:
d) \(x = \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
B27:\\
a){\left( {x + \dfrac{1}{5}} \right)^2} = 9\\
\to \left| {x + \dfrac{1}{5}} \right| = 3\\
\to \left[ \begin{array}{l}
x + \dfrac{1}{5} = 3\\
x + \dfrac{1}{5} = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{14}}{5}\\
x = - \dfrac{{16}}{5}
\end{array} \right.\\
c){\left( {2x - \dfrac{1}{3}} \right)^2} = \dfrac{9}{{25}}\\
\to \left| {2x - \dfrac{1}{3}} \right| = \dfrac{3}{5}\\
\to \left[ \begin{array}{l}
2x - \dfrac{1}{3} = \dfrac{3}{5}\\
2x - \dfrac{1}{3} = - \dfrac{3}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = \dfrac{{14}}{{15}}\\
2x = - \dfrac{4}{{15}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{{15}}\\
x = - \dfrac{2}{{15}}
\end{array} \right.\\
b)\dfrac{{22}}{9} - \dfrac{7}{3} = {\left( {x + \dfrac{1}{2}} \right)^2}\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{1}{9}\\
\to \left| {x + \dfrac{1}{2}} \right| = \dfrac{1}{9}\\
\to \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{1}{9}\\
x + \dfrac{1}{2} = - \dfrac{1}{9}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{7}{{18}}\\
x = - \dfrac{{22}}{{18}}
\end{array} \right.\\
d){\left( {\dfrac{1}{2}x + \dfrac{1}{3}} \right)^3} = \dfrac{8}{{27}}\\
\to \dfrac{1}{2}x + \dfrac{1}{3} = \sqrt[3]{{\dfrac{8}{{27}}}}\\
\to \dfrac{1}{2}x + \dfrac{1}{3} = \dfrac{2}{3}\\
\to \dfrac{1}{2}x = \dfrac{1}{3}\\
\to x = \dfrac{2}{3}
\end{array}\)
