c/đễ A<0 <=> -1/X-2 <0 <=> x-2<0 <=>x<2
a) \(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right)\)
\(A=\left(\frac{x}{x^2-4}-\frac{2\left(x+2\right)}{x^2-4}+\frac{x-2}{x^2-4}\right):\frac{6}{x+2}\)
\(A=\frac{x-2\left(x+2\right)+x-2}{x^2-4}:\frac{6}{x+2}\)
\(A=\frac{-6}{x^2-4}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
b) Ta có \(\left|x\right|=\frac{1}{2}\Leftrightarrow x=\pm\frac{1}{2}\)
TH1: Nếu \(x=\frac{1}{2}\)thì:
\(A=\frac{-1}{\frac{1}{2}-2}=\frac{-1}{\frac{-3}{2}}=-1\times\frac{2}{-3}=\frac{2}{3}\)
TH2: nếu \(x=\frac{-1}{2}\)thì:
\(A=\frac{-1}{\frac{-1}{2}-2}=\frac{-1}{\frac{-5}{2}}=-1\times\frac{2}{-5}=\frac{2}{5}\)
Vậy tại \(\left|x\right|=\frac{1}{2}\)thì \(A=\left\{\frac{2}{3};\frac{2}{5}\right\}\)
c) Để \(A< 0\)thì \(\frac{-1}{x-2}< 0\)
\(\Leftrightarrow x-2>0\)
\(\Leftrightarrow x>2\)
Vậy để \(A< 0\)thì \(x>2\)