3x^2+5x+1 trên x^3-1 – 1-x trên x^2+x+1 + 3 trên 1-x 22/08/2021 Bởi Abigail 3x^2+5x+1 trên x^3-1 – 1-x trên x^2+x+1 + 3 trên 1-x
Đáp án: $\begin{array}{l}\frac{{3{x^2} + 5x + 1}}{{{x^3} – 1}} – \frac{{1 – x}}{{{x^2} + x + 1}} + \frac{3}{{1 – x}}\\ = \frac{{3{x^2} + 5x + 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} – \frac{{1 – x}}{{{x^2} + x + 1}} – \frac{3}{{x – 1}}\\ = \frac{{3{x^2} + 5x + 1 – \left( {1 – x} \right)\left( {x – 1} \right) – 3\left( {{x^2} + x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{3{x^2} + 5x + 1 – {{\left( {x – 1} \right)}^2} – 3{x^2} – 3x – 3}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{2x – 2 – {x^2} + 2x – 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ – {x^2} + 4x – 3}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ – \left( {x – 1} \right)\left( {x – 3} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{3 – x}}{{{x^2} + x + 1}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\frac{{3{x^2} + 5x + 1}}{{{x^3} – 1}} – \frac{{1 – x}}{{{x^2} + x + 1}} + \frac{3}{{1 – x}}\\
= \frac{{3{x^2} + 5x + 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} – \frac{{1 – x}}{{{x^2} + x + 1}} – \frac{3}{{x – 1}}\\
= \frac{{3{x^2} + 5x + 1 – \left( {1 – x} \right)\left( {x – 1} \right) – 3\left( {{x^2} + x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \frac{{3{x^2} + 5x + 1 – {{\left( {x – 1} \right)}^2} – 3{x^2} – 3x – 3}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \frac{{2x – 2 – {x^2} + 2x – 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \frac{{ – {x^2} + 4x – 3}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \frac{{ – \left( {x – 1} \right)\left( {x – 3} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \frac{{3 – x}}{{{x^2} + x + 1}}
\end{array}$