Cho phương trình \({2^{\left| {\frac{{28}}{3}x + 1} \right|}} = {16^{{x^2} - 1}}\) . Khẳng định nào sau đây là đúng?

* Hướng dẫn giải

Ta có : \({2^{\left| {\dfrac{{28}}{3}x + 1} \right|}} = {16^{{x^2} - 1}} \Leftrightarrow {2^{\left| {\dfrac{{28}}{3}x + 1} \right|}} = {2^{4\left( {{x^2} - 1} \right)}}\) \( \Leftrightarrow \left| {\dfrac{{28}}{3}x + 1} \right| = 4\left( {{x^2} - 1} \right)\,\,\,\left( {DK:\,\,\left[ \begin{array}{l}x \ge 1\\x \le  - 1\end{array} \right.} \right)\) \( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\dfrac{{28}}{3}x + 1 = 4\left( {{x^2} - 1} \right){\mkern 1mu} }\\{\dfrac{{28}}{3}x + 1 = 4\left( {1 - {x^2}} \right){\mkern 1mu} }\end{array}} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}12{x^2} - 28x - 15 = 0\\12{x^2} + 28x - 9 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7 + \sqrt {94} }}{6}\,\,\,\,\,\left( {tm} \right)\\x = \frac{{7 - \sqrt {94} }}{6}\,\,\,\,\,\left( {ktm} \right)\\x = \frac{{ - 7 + 2\sqrt {19} }}{6}\,\,\,\,\left( {ktm} \right)\\x = \frac{{ - 7 - 2\sqrt {19} }}{6}\,\,\,\,\left( {tm} \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{7 + \sqrt {94} }}{6}\\x = \frac{{ - 7 - 2\sqrt {19} }}{6}\end{array} \right.\)

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