Toán Lớp 8: $\frac{5}{x+1}$ – $\frac{10}{x-x^{2}-1}$ – $\frac{15}{x^{3}+1}$ = $\frac{4}{x}$
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Toán Lớp 8: $\frac{5}{x+1}$ – $\frac{10}{x-x^{2}-1}$ – $\frac{15}{x^{3}+1}$ = $\frac{4}{x}$
Comments ( 1 )
Dkxd:\left\{ \begin{array}{l}
x + 1 \ne 0\\
{x^3} + 1 \ne 0\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne – 1\\
x \ne 0
\end{array} \right.\\
\dfrac{5}{{x + 1}} – \dfrac{{10}}{{x – {x^2} – 1}} – \dfrac{{15}}{{{x^3} + 1}} = \dfrac{4}{x}\\
\Leftrightarrow \dfrac{5}{{x + 1}} + \dfrac{{10}}{{{x^2} – x + 1}} – \dfrac{{15}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = \dfrac{4}{x}\\
\Leftrightarrow \dfrac{{5.\left( {{x^2} – x + 1} \right) + 10\left( {x + 1} \right) – 15}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = \dfrac{4}{x}\\
\Leftrightarrow \dfrac{{5{x^2} – 5x + 5 + 10x + 10 – 15}}{{{x^3} + 1}} = \dfrac{4}{x}\\
\Leftrightarrow \left( {5{x^2} + 5x} \right).x = 4\left( {{x^3} + 1} \right)\\
\Leftrightarrow 5{x^2}\left( {x + 1} \right) – 4\left( {x + 1} \right)\left( {{x^2} – x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right).\left( {5{x^2} – 4{x^2} + 4x – 4} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 4x – 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\left( {ktm} \right)\\
{x^2} + 4x – 4 = 0
\end{array} \right.\\
\Leftrightarrow {x^2} + 4x + 4 – 8 = 0\\
\Leftrightarrow {\left( {x + 2} \right)^2} = 8\\
\Leftrightarrow x = – 2 \pm 2\sqrt 2 \left( {tm} \right)\\
Vậy\,x = – 2 \pm 2\sqrt 2
\end{array}$