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$\begin{array}{l} \sqrt {9{x^2} + 6x + 1} = \sqrt {11 - 6\sqrt 2 } \\ \Leftrightarrow \sqrt {{{\left( {3x + 1} \right)}^2}} = \sqrt {9 - 2.3\sqrt 2 + {{\left( {\sqrt 2 } \right)}^2}} \\ \Leftrightarrow \left| {3x + 1} \right| = \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} \\ \Leftrightarrow \left| {3x + 1} \right| = 3 - \sqrt 2 \,\,\,\,\left( {do\,\,3 - \sqrt 2 > 0} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 3x + 1 = 3 - \sqrt 2 \\ 3x + 1 = \sqrt 2 - 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 3x = 2 - \sqrt 2 \\ 3x = \sqrt 2 - 4 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{2 - \sqrt 2 }}{3}\\ x = \frac{{\sqrt 2 - 4}}{3} \end{array} \right.. \end{array}$
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