Question

已知X^2-5X-1991=0 则[(X-2)^4+(X-1)^2-1] / (X-1)(X-2) 的值为

Answer 1

[(X-2)^4+(X-1)^2-1] / (X-1)(X-2)
={(X-2)^4+[(X-2)+1]^2-1}/(X-1)(X-2)
=[(X-2)^4+(X-2)^2+2*(X-2)+1-1]/(X-1)(X-2)
=[(X-2)^3+(X-2)+2]/(X-1)
=[(X-2)^3+X]/(X-1)
={[(X-1)-1]^3+X}/(X-1)
=[(X-1)^3-3*(X-1)^2+3*(X-1)-1+X]/(X-1)
=(X-1)^2-3(X-1)+2
=[(X-1)-1][(X-1)-2]
=(X-2)(X-3)
=X^2-5X+6
已知X^2-5X-1991=0 即：X^2-5X=1991
∴[(X-2)^4+(X-1)^2-1] / (X-1)(X-2)=X^2-5X+6=1991+6=1997

追问

=[(X-1)^3-3*(X-1)^2+3*(X-1)-1+X]/(X-1)
这块总觉得应该等于
=(X-1)^2-3(X-1)+4啊

Answer 2

[(x-2)^4+(x-1)^2-1] / (x-1)(x-2)
=[(x-2)^4+x^2-2x+1-1] / (x-1)(x-2)
=[(x-2)^4+x(x-2)]/(x-1)(x-2)
=[(x-2)^3+x]/(x-1)
=(x^3-6x^2+12x-8+x)/(x-1)
=[x(x^2-6x+5)+8x-8]/(x-1)
=[x(x-1)(x-5)+8(x-1)]/(x-1)
=x(x-5)+8
=x^2-5x+8
=1991+8
=1999