Giải giùm mình phương trình x(5$x^{3}$ + 2) – 2($\sqrt{2x + 1}$ – 1 ) = 0 04/10/2021 Bởi aikhanh Giải giùm mình phương trình x(5$x^{3}$ + 2) – 2($\sqrt{2x + 1}$ – 1 ) = 0
\[\begin{array}{l} x\left( {5{x^3} + 2} \right) – 2\left( {\sqrt {2x + 1} – 1} \right) = 0\,\,\,\left( * \right)\\ DK:\,\,\,x \ge – \frac{1}{2}.\\ \left( * \right) \Leftrightarrow 5{x^4} + 2x – 2\sqrt {2x + 1} + 2 = 0\\ \Leftrightarrow 2x + 1 – 2\sqrt {2x + 1} + 1 + 5{x^4} = 0\\ \Leftrightarrow {\left( {\sqrt {2x + 1} – 1} \right)^2} + 5{x^4} = 0\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {2x + 1} – 1 = 0\\ {x^4} = 0 \end{array} \right.\,\,\,\left( {do\,\,\,\left\{ \begin{array}{l} {\left( {\sqrt {2x + 1} – 1} \right)^2} \ge 0\,\,\forall x \ge – \frac{1}{2}\\ {x^4} \ge 0\,\,\forall x \end{array} \right.} \right)\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {2x + 1} = 1\\ x = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2x + 1 = 1\\ x = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 0\\ x = 0 \end{array} \right. \Leftrightarrow x = 0\,\,\,\left( {tm} \right).\\ Vay\,\,\,pt\,\,co\,\,nghiem\,\,x = 0. \end{array}\] Bình luận
Hiểu biết x ( 5 x 3 + 2 ) − 2 ( √ 2 x + 1 − 1 ) = 0 ( ∗ ) D K : x ≥ − 1 2 . ( ∗ ) ⇔ 5 x 4 + 2 x − 2 √ 2 x + 1 + 2 = 0 ⇔ 2 x + 1 − 2 √ 2 x + 1 + 1 + 5 x 4 = 0 ⇔ ( √ 2 x + 1 − 1 ) 2 + 5 x 4 = 0 ⇔ { √ 2 x + 1 − 1 = 0 x 4 = 0 ( d o { ( √ 2 x + 1 − 1 ) 2 ≥ 0 ∀ x ≥ − 1 2 x 4 ≥ 0 ∀ x ) ⇔ { √ 2 x + 1 = 1 x = 0 ⇔ { 2 x + 1 = 1 x = 0 ⇔ { x = 0 x = 0 ⇔ x = 0 ( t m ) . V a y p t c o n g h i e m x = 0. Bình luận
\[\begin{array}{l}
x\left( {5{x^3} + 2} \right) – 2\left( {\sqrt {2x + 1} – 1} \right) = 0\,\,\,\left( * \right)\\
DK:\,\,\,x \ge – \frac{1}{2}.\\
\left( * \right) \Leftrightarrow 5{x^4} + 2x – 2\sqrt {2x + 1} + 2 = 0\\
\Leftrightarrow 2x + 1 – 2\sqrt {2x + 1} + 1 + 5{x^4} = 0\\
\Leftrightarrow {\left( {\sqrt {2x + 1} – 1} \right)^2} + 5{x^4} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {2x + 1} – 1 = 0\\
{x^4} = 0
\end{array} \right.\,\,\,\left( {do\,\,\,\left\{ \begin{array}{l}
{\left( {\sqrt {2x + 1} – 1} \right)^2} \ge 0\,\,\forall x \ge – \frac{1}{2}\\
{x^4} \ge 0\,\,\forall x
\end{array} \right.} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {2x + 1} = 1\\
x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x + 1 = 1\\
x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
x = 0
\end{array} \right. \Leftrightarrow x = 0\,\,\,\left( {tm} \right).\\
Vay\,\,\,pt\,\,co\,\,nghiem\,\,x = 0.
\end{array}\]
Hiểu biết
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