原式=[(x²+2x)/(1+x)]÷[(x-2)/(x+1)]=[(x²+2x)/(x+1)]×[(x+1)/(x-2)]=(x²+2x)/(x-2)=[(-1/2)²+2×(-1/2)]/(-1/2-2)=(1/4-1)/(-5/2)=(-3/4)×(-2/5)=3/10是這樣的x減2分之x+1)不是這樣 [(x-2)/(x+1)]你好：是這樣嗎？原式=[(x²+2x)/(x+1)]÷[(x-2)/x+1]=[(x²+2x)/(x+1)]÷[(x-2)/x+(x/x)]=[(x²+2x)/(x+1)]÷[(x-2+x)/x]=[x(x+2)/(x+1)]×{x/[2(x-1)]=x²(x+2)/(x²-1)=(x³+2x²)/(x²-1)=[(-1/2)³+2×(-1/2)²]/[(-1/2)²-1]=(-1/8+1/2)/(-3/4)=3/8×(-4/3)=-1/2