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

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