( 2/x-1 – 1/x+1) x^2-1 /x^2- 6x +9 + x+1 /2x- 6 giúp mình với ạ 31/07/2021 Bởi Amara ( 2/x-1 – 1/x+1) x^2-1 /x^2- 6x +9 + x+1 /2x- 6 giúp mình với ạ
Đáp án: $\frac{{{x^2} + 3}}{{2{{(x – 3)}^2}}}$ Giải thích các bước giải:ĐKXĐ $\{ _{x \ne 3}^{x \ne \pm 1}$ $\begin{array}{l}\left( {\frac{2}{{x – 1}} – \frac{1}{{x + 1}}} \right)\left( {\frac{{{x^2} – 1}}{{{x^2} – 6x + 9}}} \right) + \frac{{x + 1}}{{2x – 6}}\\ = \frac{{2(x + 1) – (x – 1)}}{{(x – 1)(x + 1)}}.\frac{{{x^2} – 1}}{{{{(x – 3)}^2}}} + \frac{{x + 1}}{{2(x – 3)}}\\ = \frac{{x + 3}}{{{x^2} – 1}}.\frac{{{x^2} – 1}}{{{{(x – 3)}^2}}} + \frac{{(x + 1)(x – 3)}}{{2{{(x – 3)}^2}}}\\ = \frac{{2(x + 3)}}{{2{{(x – 3)}^2}}} + \frac{{{x^2} – 2x – 3}}{{2{{(x – 3)}^2}}}\\ = \frac{{{x^2} + 3}}{{2{{(x – 3)}^2}}}\end{array}$ Bình luận
Đáp án:
$\frac{{{x^2} + 3}}{{2{{(x – 3)}^2}}}$
Giải thích các bước giải:ĐKXĐ
$\{ _{x \ne 3}^{x \ne \pm 1}$
$\begin{array}{l}
\left( {\frac{2}{{x – 1}} – \frac{1}{{x + 1}}} \right)\left( {\frac{{{x^2} – 1}}{{{x^2} – 6x + 9}}} \right) + \frac{{x + 1}}{{2x – 6}}\\
= \frac{{2(x + 1) – (x – 1)}}{{(x – 1)(x + 1)}}.\frac{{{x^2} – 1}}{{{{(x – 3)}^2}}} + \frac{{x + 1}}{{2(x – 3)}}\\
= \frac{{x + 3}}{{{x^2} – 1}}.\frac{{{x^2} – 1}}{{{{(x – 3)}^2}}} + \frac{{(x + 1)(x – 3)}}{{2{{(x – 3)}^2}}}\\
= \frac{{2(x + 3)}}{{2{{(x – 3)}^2}}} + \frac{{{x^2} – 2x – 3}}{{2{{(x – 3)}^2}}}\\
= \frac{{{x^2} + 3}}{{2{{(x – 3)}^2}}}
\end{array}$
Đáp án:
Giải thích các bước giải: