\(A=\left(\frac{2x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{5-x^2}{x+2}\right)\) ĐKXĐ : \(x\ne\pm2\)
\(A=\left(\frac{2x}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4}{x+2}+\frac{5-x^2}{x+2}\right)\)
\(A=\left(\frac{2x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+5-x^2}{x+2}\right)\)
\(A=\frac{x-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1}\)
\(A=\frac{x-6}{x-2}\)
b, ta có \(/\frac{1}{2}/=\frac{1}{2}=\frac{-1}{2}\)
TH1 : Thay x = 1/2 vào A
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Th2 : Thay x = -1/2 vào A :
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Bn tự tính vào kết luận
c, Để \(A< 0\) \(\Rightarrow\frac{x-6}{x-2}\)\(< 0\)
Trường hợp 1 : \(\hept{\begin{cases}x-6>0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>6\\x< 2\end{cases}\Rightarrow x\in}\varnothing}\)
Trường hợp 2 \(\hept{\begin{cases}x-6< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< 6\\x>2\end{cases}\Rightarrow}2< x< 6}\)
Vậy để \(A< 0\)thì \(2< x< 6\)