∵2x²-2x-3=0,∴2x²-1=2x+2=2(x+1)∴(x-1)²/(x²-1)+ (2x²-x)/(x+1)=(x-1)²/[(x-1)(x+1)]+(2x²-x)/(x+1)=（x-1)/(x+1)+(2x²-x)/(x+1)=(2x²-1)/(x+1)=2(x+1)/(x+1)=2...謝謝請(qǐng)問(wèn)還能再回答我一個(gè)問(wèn)題嗎？簡(jiǎn)單的先化簡(jiǎn)（1-a-1分之1）除以a²-a分之a(chǎn)²-4a+4再在1.0.-1中選擇一個(gè)作為a的值，求出原方程式的值？您看行么？真是謝謝了[1-1/(a-1)]÷[(a²-4a+4)/(a²-a)]=[(a-1)-1]/(a-1)÷{(a-2)²/[a(a-1)]}=(a-2)/(a-1)×[a(a-1)/(a-2)²]=a/(a-2)原式有意義需分母不等于0那么a²-a≠0,,a-1≠0 ∴a≠0且a≠1在1.0.-1中a只可取-1當(dāng)a=-1時(shí)，原式=1/3