有理分式拆分用待定系数法,书上都有的。例:设1/(x^2+2x-8)=1/[(x-2)(x+4)]=A/(x-2)+B/(x+4)=[(A+B)x+(4A-2B)]/[(x-2)(x+4)],则A+B=0;
4A-2B=1,联立解得A=1/6,B=-1/6得1/(x^2+2x-8)=(1/6)[1/(x-2)-1/(x+4)]
有理分式的拆分公式急求答案,帮忙回答下
有理分式拆分用待定系数法,书上都有的。例:设1/(x^2+2x-8)=1/[(x-2)(x+4)]=A/(x-2)+B/(x+4)=[(A+B)x+(4A-2B)]/[(x-2)(x+4)],则A+B=0;
4A-2B=1,联立解得A=1/6,B=-1/6得1/(x^2+2x-8)=(1/6)[1/(x-2)-1/(x+4)]