1)设x+5=y1/(y+1)+1/(y-1)=1/(y+2)+1/(y-2)2y/(y+1)(y-1)=2y/(y+2)(y-2)y(y+2)(y-2)=y(y+1)(y-1)y(y^2-4)=y（y^2-1)y=0x+5=0x=-52)设x-5=y1/(y+2)+1/(y-2)=1/(y+1)+1/(y-1)2y/(y+2)(y-2)=2y/(y+1)(y-1)y(y+1)(y-1)=y...不是这样算的吧，(1)应该化成:1/(x+6)+1/(x+7)=1/(x+4)+1/(x+3)(2)应该化成:1/(x-3)+1/(x-4)=1/(x-7)+1/(x-6)你可以试试，没我的算法简单呦