Tìm \(x\), biết:
a) \(\displaystyle {\left( {2x + 3} \right)^2} = {9 \over {121}}\);
b) \(\displaystyle {\left( {3x - 1} \right)^3} = - {8 \over {27}}\)
Hướng dẫn giải
\(\begin{array}{l}
{A^2} = {B^2} \Rightarrow \left[ \begin{array}{l}
A = B\\
A = - B
\end{array} \right.\\
{A^3} = {B^3} \Rightarrow A = B
\end{array}\)
Lời giải chi tiết
a) \(\displaystyle {\left( {2x + 3} \right)^2} = {9 \over {121}}\)
\({\left( {2x + 3} \right)^2} = \dfrac{{{3^2}}}{{{{11}^2}}}\)
\(\displaystyle {\left( {2x + 3} \right)^2} = {\left( { {3 \over {11}}} \right)^2}\)
\( \Rightarrow \left[ \begin{array}{l}
2x + 3 = \dfrac{3}{{11}}\\
2x + 3 = \dfrac{{ - 3}}{{11}}
\end{array} \right.\)
\(\begin{array}{l}
+ )\,\,2x + 3 = \dfrac{3}{{11}}\\
\,\,\,\,\,\,\;2x = \dfrac{3}{{11}} - 3\\
\,\,\,\,\,\,\;2x = \dfrac{3}{{11}} - \dfrac{{33}}{{11}}\\
\,\,\,\,\,\,\;2x = \dfrac{{ - 30}}{{11}}\\
\,\,\,\,\,\,\,\,\,x = \dfrac{{ - 30}}{{11}}:2\\
\,\,\,\,\,\,\,\,\,x = \dfrac{{ - 30}}{{11}}.\dfrac{1}{2} = \dfrac{{ - 15}}{{11}}\\
+ )\,\,2x + 3 = \dfrac{{ - 3}}{{11}}\\
\,\,\,\,\,\,\;2x = \dfrac{{ - 3}}{{11}} - 3\\
\,\,\,\,\,\,\;2x = \dfrac{{ - 3}}{{11}} + \dfrac{{ - 33}}{{11}}\\
\,\,\,\,\,\,\;2x = \dfrac{{ - 36}}{{11}}\\
\,\,\,\,\,\,\,\,\,x = \dfrac{{ - 36}}{{11}}:2\\
\,\,\,\,\,\,\,\,\,x = \dfrac{{ - 36}}{{11}}.\dfrac{1}{2} = \dfrac{{ - 18}}{{11}}
\end{array}\)
Vậy \(x=\dfrac{{ - 18}}{{11}}\) hoặc \(x=\dfrac{{ - 15}}{{11}}\)
b) \(\displaystyle {\left( {3x - 1} \right)^3} = - {8 \over {27}} = {\left( { - {2 \over 3}} \right)^3} \)
\(\displaystyle \Rightarrow 3x - 1 = - {2 \over 3} \)
\(\begin{array}{l}
\Rightarrow 3x = \dfrac{{ - 2}}{3} + 1\\
\Rightarrow 3x = \dfrac{{ - 2}}{3} + \dfrac{3}{3}\\
\Rightarrow 3x = \dfrac{1}{3}\\
\Rightarrow x = \dfrac{1}{3}:3 = \dfrac{1}{9}
\end{array}\)
Vậy \(x=\dfrac{1}{9}\)
-- Mod Toán 7