a) \(A=B\) khi
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=-16\)
\(\Leftrightarrow8x=-16\)
\(\Leftrightarrow x=\dfrac{-16}{8}\)
\(\Leftrightarrow x=-2\left(ktmdk\right)\)
b) \(A:B< 0\) khi:
\(\left(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}\right):\left(\dfrac{-16}{x^2-4}\right)< 0\)
\(\Leftrightarrow\left[\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\right]:\left[\dfrac{-16}{\left(x+2\right)\left(x-2\right)}\right]< 0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-16}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{x}{-2}< 0\)
Mà: -2 < 0
\(\Leftrightarrow x>0\)
So với đk:
Vậy: \(A:B< 0\) khi
\(x>0;x\ne2\)
a: A=B
=>A-B=0
=>\(\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
=>x^2+4x+4-x^2+4x-4=-16
=>8x=-16
=>x=-2(loại)
b: A:B<0
=>\(\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
=>\(\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
=>\(\dfrac{-8x}{16}< 0\)
=>x>0
Kết hợp ĐKXĐ, ta được: x>0 và x<>2