Đáp án:
`a)A=(x/(x^2-4)+2/(2-x)+1/(x+2)):(x-2+(10-x^2)/(x+2))`
`đk:x ne +-2`
`A=(x/(x^2-4)-2/(x-2)+1/(x+2)):((x^2-4+10-x^2)/(x+2))`
`=((x-2(x+2)+x+2)/(x^2-4)):6/(x+2)`
`=(-2/(x^2-4))*(x+2)/6`
`=-1/(3(x-2))`
`b)|x|=1/2`
`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=-\dfrac12\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}A=\dfrac{-1}{3\Big(\dfrac12-2\Big)}=\dfrac29\\A=\dfrac{-1}{3\Big(\dfrac{-1}{2}-2\Big)}=\dfrac{2}{15}\end{array} \right.\)
`c)A<0`
`<=>(-1)/(3(x-2))<0`
`<=>1/(3(x-2))>0`
`<=>x-2>0` do `1/3>0`
`<=>x>2`.
