a, \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right)\frac{x^2+4x+4}{8}\)ĐK : \(x\ne\pm2\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+2-2x+2}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}=\frac{4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{2\left(x-2\right)}=\frac{x+2}{2x-4}\)
b, A = x hay
\(\frac{x+2}{2x-4}=x\Leftrightarrow x+2=2x^2-4x\)
\(\Leftrightarrow5x+2-2x^2=0\)vô nghiệm
tương tự với A = x/2 nhé !
