y=a/c+(bc-ad)/(c²) *1/(x+(d/c))=a/c+ (bc-ad)/[c(cx+d)]=[a(cx+d) + (bc-ad)]/[c(cx+d)]=[acx+ad)+ (bc-ad)]/[c(cx+d)]=(ax+b)/(cx+d)x+3+8/(x-2)-1/(x-1)=x+3+[8(x-1)-(x-2)]/( (x-2)(x-1) )=(x+3)+(7x-6)/(...朋友 不是让你反向推，不是让你同分的。是让你分解的。(ax+b)/(cx+d)=[acx+ad)+(bc-ad)]/[c(cx+d)]=[a(cx+d)+ (bc-ad)]/[c(cx+d)]=a/c+(bc-ad)/[c(cx+d)]=a/c+(bc-ad)/(c²)*1/(x+(d/c))(x+3)+(7x-6)/( (x-2)(x-1) )=x+3+[8(x-1)-(x-2)]/((x-2)(x-1) )=x+3+8/(x-2)-1/(x-1)x³/(x²-3x+2)=[x^3-3x^2+2x+3x^2-9x+6+7x-6]/(x²-3x+2)=[(x+3)(x^2-3x+2)+7x-6]/(x²-3x+2)=[(x+3)(x-2)(x-1)+(7x-6)]/((x-2)(x-1) )=(x+3)+(7x-6)/((x-2)(x-1) )