cộng, trừ các phân thức sau: x/2(x-1) + x-2/x^2 – 1 -5/2x+2 16/08/2021 Bởi Nevaeh cộng, trừ các phân thức sau: x/2(x-1) + x-2/x^2 – 1 -5/2x+2
Đáp án: $\begin{array}{l}\frac{x}{{2\left( {x – 1} \right)}} + \frac{{x – 2}}{{{x^2} – 1}} – \frac{5}{{2x + 2}}\left( {dkxd:x \ne \pm 1} \right)\\ = \frac{x}{{2\left( {x – 1} \right)}} + \frac{{x – 2}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} – \frac{5}{{2\left( {x + 1} \right)}}\\ = \frac{{x\left( {x + 1} \right) + 2\left( {x – 2} \right) – 5\left( {x – 1} \right)}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \frac{{{x^2} + x + 2x – 4 – 5x + 5}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \frac{{{x^2} – 2x + 1}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \frac{{{{\left( {x – 1} \right)}^2}}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \frac{{x – 1}}{{2\left( {x + 1} \right)}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\frac{x}{{2\left( {x – 1} \right)}} + \frac{{x – 2}}{{{x^2} – 1}} – \frac{5}{{2x + 2}}\left( {dkxd:x \ne \pm 1} \right)\\
= \frac{x}{{2\left( {x – 1} \right)}} + \frac{{x – 2}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} – \frac{5}{{2\left( {x + 1} \right)}}\\
= \frac{{x\left( {x + 1} \right) + 2\left( {x – 2} \right) – 5\left( {x – 1} \right)}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^2} + x + 2x – 4 – 5x + 5}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^2} – 2x + 1}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{{\left( {x – 1} \right)}^2}}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{x – 1}}{{2\left( {x + 1} \right)}}
\end{array}$