Solution Let: 12x2−48(x−2)2(x+2)2=Ax−2+B(x−2)2+Cx+2+D(x+2)2\frac{12x^2-48}{(x-2)^2(x+2)^2}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+2}+\frac{D}{(x+2)^2}(x−2)2(x+2)212x2−48=x−2A+(x−2)2B+x+2C+(x+2)2D Solving gives: A=3,B=0,C=−3,D=0A=3,\quad B=0,\quad C=-3,\quad D=0A=3,B=0,C=−3,D=0 Hence: 12x2−48(x−2)2(x+2)2=3x−2−3x+2\boxed{\frac{12x^2-48}{(x-2)^2(x+2)^2}=\frac{3}{x-2}-\frac{3}{x+2}}(x−2)2(x+2)212x2−48=x−23−x+23