Cho biểu thức A=(x/(x^2-1)+2/(2-x)+1/(x+2):(x-2+(10-x^2)/(x+2)

* Hướng dẫn giải

$\begin{array}{l}a)A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\left(x\ne \pm 2\right)\\ =\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{\left(x+2\right)\left(x-2\right)+10-x^2}{x+2}\\ =\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{x^2-4+10-x^2}\\ =\frac{-6}{x-2}.\frac{1}{6}=\frac{1}{2-x}\\ \end{array}$

$\begin{array}{l}b)\left|x\right|=\frac{1}{2}\Rightarrow \left[\begin{array}{l}x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\\ x=-\frac{1}{2}\Rightarrow A=\frac{1}{2-\left(-\frac{1}{2}\right)}=\frac{2}{5}\end{array}\right.\\ c)A<0\iff \frac{1}{2-x}<0\iff 2-x<0\left(do1>0\right)\Rightarrow x>2(tm)\end{array}$