Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {5{y^4} - 3{y^3}} \right):2{y^3} = \dfrac{1}{2}\\
\Leftrightarrow 5{y^4}:2{y^3} - 3{y^3}:2{y^3} = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{5}{2}y - \dfrac{3}{2} = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{5}{2}y = \dfrac{3}{2} + \dfrac{1}{2}\\
\Leftrightarrow \dfrac{5}{2}y = 2\\
\Leftrightarrow y = 2:\dfrac{5}{2}\\
\Leftrightarrow y = \dfrac{4}{5}\\
b,\\
\left( {{y^4} - 2{y^2} - 8} \right):\left( {y - 2} \right) = 0\\
\Leftrightarrow \left[ {\left( {{y^4} - 4{y^2}} \right) + \left( {2{y^2} - 8} \right)} \right]:\left( {y - 2} \right) = 0\\
\Leftrightarrow \left[ {{y^2}\left( {{y^2} - 4} \right) + 2.\left( {{y^2} - y} \right)} \right]:\left( {y - 2} \right) = 0\\
\Leftrightarrow \left[ {\left( {{y^2} - 4} \right)\left( {{y^2} + 2} \right)} \right]:\left( {y - 2} \right) = 0\\
\Leftrightarrow \left[ {\left( {y - 2} \right)\left( {y + 2} \right)\left( {{y^2} + 2} \right)} \right]:\left( {y - 2} \right) = 0\\
\Leftrightarrow \left( {y + 2} \right)\left( {{y^2} + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y + 2 = 0\\
{y^2} + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = - 2\\
{y^2} = - 2\,\,\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow y = - 2\\
c,\\
\left( {3{y^3} + {y^2} - 13y + 5} \right):\left( {{y^2} + 2y - 1} \right) = 10\\
\Leftrightarrow \left[ {\left( {3{y^3} - 5{y^2}} \right) + \left( {6{y^2} - 10y} \right) - \left( {3y - 5} \right)} \right]:\left( {{y^2} + 2y - 1} \right) = 10\\
\Leftrightarrow \left[ {{y^2}\left( {2y - 5} \right) + 2y\left( {2y - 5} \right) - \left( {3y - 5} \right)} \right]:\left( {{y^2} + 2y - 1} \right) = 10\\
\Leftrightarrow \left[ {\left( {2y - 5} \right)\left( {{y^2} + 2y - 1} \right)} \right]:\left( {{y^2} + 2y - 1} \right) = 10\\
\Leftrightarrow 2y - 5 = 10\\
\Leftrightarrow y = \dfrac{{15}}{2}
\end{array}\)
