Đáp án:
b) \(\dfrac{{x - 2}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
b)P = \left[ {\dfrac{{x + 2}}{{2\left( {x - 2} \right)}} + \dfrac{{x - 2}}{{2\left( {x + 2} \right)}} - \dfrac{8}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right]:\dfrac{4}{{x - 2}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2} + {{\left( {x - 2} \right)}^2} - 16}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x - 2}}{4}\\
= \dfrac{{{x^2} + 4x + 4 + {x^2} - 4x + 4 - 16}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x - 2}}{4}\\
= \dfrac{{2{x^2} - 8}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x - 2}}{4}\\
= \dfrac{{2\left( {x - 2} \right)\left( {x + 2} \right)}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x - 2}}{4}\\
= \dfrac{{x - 2}}{4}\\
c)Thay:x = - \dfrac{1}{3}\\
\to P = \dfrac{{ - \dfrac{1}{3} - 2}}{4} = - \dfrac{7}{{12}}
\end{array}\)
