((x+1)/(x-1)-(x-1)/(x+1)):(2/(x^2-1)-x/(x-1)+1/(x+1)) giúp vs 07/11/2021 Bởi Katherine ((x+1)/(x-1)-(x-1)/(x+1)):(2/(x^2-1)-x/(x-1)+1/(x+1)) giúp vs
Đáp án: \(\dfrac{{4x}}{{ – {x^2} + 1}}\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne \pm 1\\\left( {\dfrac{{x + 1}}{{x – 1}} – \dfrac{{x – 1}}{{x + 1}}} \right):\left( {\dfrac{2}{{{x^2} – 1}} – \dfrac{x}{{x – 1}} + \dfrac{1}{{x + 1}}} \right)\\ = \dfrac{{{x^2} + 2x + 1 – {x^2} + 2x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}:\dfrac{{2 – x\left( {x + 1} \right) + x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \dfrac{{4x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{ – {x^2} + 1}}\\ = \dfrac{{4x}}{{ – {x^2} + 1}}\end{array}\) Bình luận
Đáp án:
\(\dfrac{{4x}}{{ – {x^2} + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \pm 1\\
\left( {\dfrac{{x + 1}}{{x – 1}} – \dfrac{{x – 1}}{{x + 1}}} \right):\left( {\dfrac{2}{{{x^2} – 1}} – \dfrac{x}{{x – 1}} + \dfrac{1}{{x + 1}}} \right)\\
= \dfrac{{{x^2} + 2x + 1 – {x^2} + 2x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}:\dfrac{{2 – x\left( {x + 1} \right) + x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{4x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{ – {x^2} + 1}}\\
= \dfrac{{4x}}{{ – {x^2} + 1}}
\end{array}\)