Giải giùm mình phương trình x(5$x^{3}$ + 2) - 2($\sqrt{2x + 1}$ - 1 ) = 0
2 câu trả lời
\[\begin{array}{l} x\left( {5{x^3} + 2} \right) - 2\left( {\sqrt {2x + 1} - 1} \right) = 0\,\,\,\left( * \right)\\ DK:\,\,\,x \ge - \frac{1}{2}.\\ \left( * \right) \Leftrightarrow 5{x^4} + 2x - 2\sqrt {2x + 1} + 2 = 0\\ \Leftrightarrow 2x + 1 - 2\sqrt {2x + 1} + 1 + 5{x^4} = 0\\ \Leftrightarrow {\left( {\sqrt {2x + 1} - 1} \right)^2} + 5{x^4} = 0\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {2x + 1} - 1 = 0\\ {x^4} = 0 \end{array} \right.\,\,\,\left( {do\,\,\,\left\{ \begin{array}{l} {\left( {\sqrt {2x + 1} - 1} \right)^2} \ge 0\,\,\forall x \ge - \frac{1}{2}\\ {x^4} \ge 0\,\,\forall x \end{array} \right.} \right)\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {2x + 1} = 1\\ x = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2x + 1 = 1\\ x = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 0\\ x = 0 \end{array} \right. \Leftrightarrow x = 0\,\,\,\left( {tm} \right).\\ Vay\,\,\,pt\,\,co\,\,nghiem\,\,x = 0. \end{array}\]
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