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Đáp án:
$\begin{array}{l}
a){\left( {5x - \dfrac{1}{3}} \right)^2} = \dfrac{4}{9}\\
\Leftrightarrow {\left( {5x - \dfrac{1}{3}} \right)^2} = {\left( {\dfrac{2}{3}} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
5x - \dfrac{1}{3} = \dfrac{2}{3}\\
5x - \dfrac{1}{3} = - \dfrac{2}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x = \dfrac{2}{3} + \dfrac{1}{3} = 1\\
5x = - \dfrac{2}{3} + \dfrac{1}{3} = - \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{5}\\
x = - \dfrac{1}{{15}}
\end{array} \right.\\
Vậy\,x = - \dfrac{1}{{15}};x = \dfrac{1}{5}\\
b){\left( {\dfrac{2}{3} + x} \right)^3} + 27 = 0\\
\Leftrightarrow {\left( {\dfrac{2}{3} + x} \right)^3} = - 27\\
\Leftrightarrow {\left( {\dfrac{2}{3} + x} \right)^3} = {\left( { - 3} \right)^3}\\
\Leftrightarrow \dfrac{2}{3} + x = - 3\\
\Leftrightarrow x = - 3 - \dfrac{2}{3}\\
\Leftrightarrow x = - \dfrac{{11}}{3}\\
Vậy\,x = - \dfrac{{11}}{3}\\
c)\dfrac{{2x - 1}}{2} = \dfrac{8}{{2x - 1}}\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = 2.8\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = 16\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 4\\
2x - 1 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 4 + 1\\
2x = - 4 + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 5\\
2x = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = - \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{2};x = - \dfrac{3}{2}\\
d){3^{2x}} + {3^{2x + 1}} = 324\\
\Leftrightarrow {3^{2x}} + {3^{2x}}.3 = 324\\
\Leftrightarrow {3^{2x}}.\left( {1 + 3} \right) = 324\\
\Leftrightarrow {3^{2x}}.4 = 324\\
\Leftrightarrow {3^{2x}} = 81\\
\Leftrightarrow {3^{2x}} = {3^4}\\
\Leftrightarrow 2x = 4\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
e)Dkxd:x \ge 0\\
\Leftrightarrow \sqrt x = 9\\
\Leftrightarrow x = {9^2}\\
\Leftrightarrow x = 81\left( {tm} \right)\\
Vậy\,x = 81\\
f)Dkxd:2x - 3 \ge 0\\
\Leftrightarrow x \ge \dfrac{3}{2}\\
\sqrt {2x - 3} = 5\\
\Leftrightarrow 2x - 3 = 25\\
\Leftrightarrow 2x = 28\\
\Leftrightarrow x = 14\left( {tm} \right)\\
Vậy\,x = 14
\end{array}$
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