Đáp án:
B2:
b) \(\left[ \begin{array}{l}
x = \dfrac{1}{6}\\
x = - \dfrac{7}{6}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{4}{9} + \dfrac{4}{3} + 4.1\\
= \dfrac{{52}}{9}\\
b)B = \left( {\dfrac{1}{{16}} + \dfrac{1}{2} - \dfrac{1}{{16}}} \right):\left( {\dfrac{1}{8} - 1 + 1} \right)\\
= \dfrac{1}{2}:\dfrac{1}{8} = 4\\
B2\\
a)\left| {x - \dfrac{1}{2}} \right| = \dfrac{3}{4}\\
\to \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{3}{4}\\
x - \dfrac{1}{2} = - \dfrac{3}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{4}\\
x = - \dfrac{1}{4}
\end{array} \right.\\
b){\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{23}}{9} = \dfrac{{2,7}}{{100}}:\dfrac{{0,9}}{{100}}\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{23}}{9} = 3\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{4}{9}\\
\to \left| {x + \dfrac{1}{2}} \right| = \dfrac{2}{3}\\
\to \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{2}{3}\\
x + \dfrac{1}{2} = - \dfrac{2}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{6}\\
x = - \dfrac{7}{6}
\end{array} \right.
\end{array}\)
