Question

用简便方法计算(2010的3次方+2010的2次方-2011)分之(2010的3次方-2×2010的2次方-2008)

Answer 1

设A=2010

∴原式=(2010^3+2010^2-2011)/(2010^3-2*2010^2-2008)

=(A^3+A^2-A-1)/(A^3-2A^2-A+2)

=(A+1)(A^2-1)/[A^3-A^2-(A^2+A-2)]

=(A+1)^2(A-1)/[A^2(A-1)-(A-1)(A+2)]

=(A+1)^2(A-1)/(A-1)(A^2-A-2)

=(A+1)^2(A-1)/(A-1)(A-2)(A+1)

=(A+1)/(A-2)

=(2010+1)/(2010-2)

=2011/2008

遇到较大或较复杂的数或式子最好用换元的思想