[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)]=[1/(x-1)²-1/(x+1)²]÷[1/(x+1）+1/(x-1)]=[1/(x+1）+1/(x-1)][1/(x-1）+1/(x+1)]÷[1/(x+1）+1/(x-1)]=1/(x-1）-1/(x+1)=(x+1)/(x+1...