a.
\(A=B\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)
\(\Leftrightarrow8x+16=0\)
\(\Leftrightarrow8\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn A=B
b.
\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{16}>0\)
\(\Leftrightarrow\dfrac{x}{2}>0\)
\(\Leftrightarrow x>0\)
