Toán Lớp 8: 2x^4 + 3x^3 -3x^2 +3x +2 = 0.tìm x
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Toán Lớp 8: 2x^4 + 3x^3 -3x^2 +3x +2 = 0.tìm x
Comments ( 1 )
+ Khi:x = 0\\
\Leftrightarrow 2 = 0\left( {ktm} \right)\\
+ Khi:x \ne 0\\
2{x^4} + 3{x^3} – 3{x^2} + 3x + 2 = 0\\
\Leftrightarrow \dfrac{{2{x^4} + 3{x^3} – 3{x^2} + 3x + 2}}{{{x^2}}} = 0\\
\Leftrightarrow 2{x^2} + 3x – 3 + \dfrac{3}{x} + \dfrac{2}{{{x^2}}} = 0\\
\Leftrightarrow 2.\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) + 3.\left( {x + \dfrac{1}{x}} \right) – 3 = 0\\
\Leftrightarrow 2.\left( {{x^2} + 2.{x^2}.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^2}}}} \right) – 2.2.{x^2}.\dfrac{1}{{{x^2}}}\\
+ 3.\left( {x + \dfrac{1}{x}} \right) – 3 = 0\\
\Leftrightarrow 2.{\left( {x + \dfrac{1}{x}} \right)^2} – 2 + 3\left( {x + \dfrac{1}{x}} \right) – 3 = 0\\
\Leftrightarrow 2.{\left( {x + \dfrac{1}{x}} \right)^2} + 3\left( {x + \dfrac{1}{x}} \right) – 5 = 0\\
Dat:\left( {x + \dfrac{1}{x}} \right) = a\\
\Leftrightarrow 2{a^2} + 3a – 5 = 0\\
\Leftrightarrow 2{a^2} + 5a – 2a – 5 = 0\\
\Leftrightarrow \left( {2a + 5} \right)\left( {a – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = – \dfrac{5}{2}\\
a = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{x} = \dfrac{{ – 5}}{2}\\
x + \dfrac{1}{x} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 1 = – \dfrac{5}{2}x\\
{x^2} + 1 = x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} + 5x + 2 = 0\\
{x^2} – x + 1 = \left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow 2{x^2} + 4x + x + 2 = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 2\left( {tm} \right)\\
x = – \dfrac{1}{2}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = – 2;x = \dfrac{{ – 1}}{2}
\end{array}$