是不是[3/(x-1)-x-1]÷[(x-2)/(x²-2x+1)][3/(x-1)-x-1]÷[(x-2)/(x²-2x+1)]={[3-(x²-1)]/(x-1)}÷[(x-2)/(x-1)²]=[(4-x²)/(x-1)]÷[(x-2)/(x-1)²]=(-2-x)(x-1)=-x²-x+1