tìm x biết
a, ( 2x/3+ 1/2 )^4 = 1/81
b, ( 3/4 – 3x/5 )^3 = 8/25
c, 1/3 : ( x/2 + 3/4 ) ^2 = 4/3
d, ( 5/6 – 3x/2 )^5 = 243/1024
tìm x biết
a, ( 2x/3+ 1/2 )^4 = 1/81
b, ( 3/4 – 3x/5 )^3 = 8/25
c, 1/3 : ( x/2 + 3/4 ) ^2 = 4/3
d, ( 5/6 – 3x/2 )^5 = 243/1024
Đáp án:
$\begin{array}{l}
a){\left( {\dfrac{{2x}}{3} + \dfrac{1}{2}} \right)^4} = \dfrac{1}{{81}}\\
\Rightarrow {\left( {\dfrac{{2x}}{3} + \dfrac{1}{2}} \right)^4} = {\left( {\dfrac{1}{3}} \right)^4} = {\left( { – \dfrac{1}{3}} \right)^4}\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{2x}}{3} + \dfrac{1}{2} = \dfrac{1}{3}\\
\dfrac{{2x}}{3} + \dfrac{1}{2} = – \dfrac{1}{3}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{2x}}{3} = \dfrac{{ – 1}}{6}\\
\dfrac{{2x}}{3} = – \dfrac{5}{6}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ – 1}}{4}\\
x = – \dfrac{5}{4}
\end{array} \right.\\
b){\left( {\dfrac{3}{4} – \dfrac{{3x}}{5}} \right)^3} = \dfrac{8}{{125}}\\
\Rightarrow \dfrac{3}{4} – \dfrac{{3x}}{5} = \dfrac{2}{5}\\
\Rightarrow \dfrac{{3x}}{5} = \dfrac{3}{4} – \dfrac{2}{5} = \dfrac{7}{{20}}\\
\Rightarrow x = \dfrac{7}{{20}}.\dfrac{5}{3}\\
\Rightarrow x = \dfrac{7}{{12}}\\
Vay\,x = \dfrac{7}{{12}}\\
c)\dfrac{1}{3}:{\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^2} = \dfrac{4}{3}\\
\Rightarrow {\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^2} = \dfrac{1}{3}:\dfrac{4}{3} = \dfrac{1}{4}\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{x}{2} + \dfrac{3}{4} = \dfrac{1}{2}\\
\dfrac{x}{2} + \dfrac{3}{4} = – \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = – \dfrac{1}{4}\\
\dfrac{x}{2} = \dfrac{{ – 5}}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ – 1}}{2}\\
x = – \dfrac{5}{2}
\end{array} \right.\\
d){\left( {\dfrac{5}{6} – \dfrac{{3x}}{2}} \right)^5} = \dfrac{{243}}{{1024}}\\
\Rightarrow \dfrac{5}{6} – \dfrac{{3x}}{2} = \dfrac{3}{4}\\
\Rightarrow \dfrac{{3x}}{2} = \dfrac{5}{6} – \dfrac{3}{4}\\
\Rightarrow \dfrac{{3x}}{2} = \dfrac{1}{{12}}\\
\Rightarrow x = \dfrac{1}{{18}}\\
Vay\,x = \dfrac{1}{{18}}
\end{array}$