\(a,\)\(\left(đk:x\ne\pm2\right)\)
\(=\left(\dfrac{2}{x+2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\left(\dfrac{2\left(x+2\right)-4}{\left(x+2\right)^2}\right):\left(\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\left(\dfrac{2x+4-4}{\left(x+2\right)^2}\right).\left(-\dfrac{\left(x-2\right)\left(x+2\right)}{x}\right)\)
\(=\dfrac{2x}{\left(x+2\right)^2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=-\dfrac{x-2}{x+2}\)
\(=\dfrac{2-x}{x+2}\)
Vậy \(A=\dfrac{2-x}{x+2}\)
\(b,\) Để \(A\le-2\) thì \(x\) là :
\(\Rightarrow\dfrac{2-x}{x+2}\le-2\)
\(\Rightarrow\dfrac{2-x}{x+2}+2\le0\)
\(\Rightarrow\dfrac{2-x+2\left(x+2\right)}{x+2}\le0\)
\(\Rightarrow2-x+2x+4\le0\)
\(\Rightarrow x+6\le0\)
\(\Rightarrow x\le-6\)
Vậy \(x\le-6\) thì \(A\le-2\)
giải giúp mình câu b) được không
