làm tính chia:(2x^4+x^3-3x^2+5x-2):2x-2 trên x ;(x khác 0;x khác 1;x khác 3 )

Đáp án:

\(\dfrac{{x\left( {2x – 1} \right)\left( {x + 2} \right)\left( {{x^2} – x + 1} \right)}}{{2\left( {x – 1} \right)}}\)

Giải thích các bước giải:

\(\begin{array}{l}
\left( {2{x^4} + {x^3} – 3{x^2} + 5x – 2} \right):\dfrac{{2x – 2}}{x}\\
 = \left( {2{x^4} – {x^3} + 2{x^3} – {x^2} – 2{x^2} + x + 4x – 2} \right).\dfrac{x}{{2\left( {x – 1} \right)}}\\
 = \left[ {{x^3}\left( {2x – 1} \right) + {x^2}\left( {2x – 1} \right) – x\left( {2x – 1} \right) + 2\left( {2x – 1} \right)} \right].\dfrac{x}{{2\left( {x – 1} \right)}}\\
 = \dfrac{{x\left( {2x – 1} \right)\left( {{x^3} + x – x + 2} \right)}}{{2\left( {x – 1} \right)}}\\
 = \dfrac{{x\left( {2x – 1} \right)\left( {{x^3} + 2{x^2} – {x^2} – 2x + x + 2} \right)}}{{2\left( {x – 1} \right)}}\\
 = \dfrac{{x\left( {2x – 1} \right)\left( {{x^2}\left( {x + 2} \right) – x\left( {x + 2} \right) + \left( {x + 2} \right)} \right)}}{{2\left( {x – 1} \right)}}\\
 = \dfrac{{x\left( {2x – 1} \right)\left( {x + 2} \right)\left( {{x^2} – x + 1} \right)}}{{2\left( {x – 1} \right)}}
\end{array}\)