已知X^2-5X-1999=0则分式{(X-2)^3-(X-1)^2+1}/X-2

来源:学生作业帮助网 编辑:作业帮 时间:2024/09/28 15:24:22
已知X^2-5X-1999=0则分式{(X-2)^3-(X-1)^2+1}/X-2
x){}K#tM#t ---m v|tO=Fqƺ@I8#mZ}MR>: l(iϴ u>Іbғ݋.+4 ,d$lmT@f$TH7n\DmµFƶ F 1B

已知X^2-5X-1999=0则分式{(X-2)^3-(X-1)^2+1}/X-2
已知X^2-5X-1999=0则分式{(X-2)^3-(X-1)^2+1}/X-2

已知X^2-5X-1999=0则分式{(X-2)^3-(X-1)^2+1}/X-2
先化简分式,再将X^2-5X-1999=0代入.
{(X-2)^3-(X-1)^2+1}/(X-2)={(X-2)^3-(X^2-2X+1)+1}/(X-2)= {(X-2)^3-(X^2-2X)}/(X-2)={(X-2)^3-X(X-2)}/(X-2)=(X-2){(X-2)^2-X}/(X-2)= (X-2)^2-X=X^2-5X+4= (X^2-5X-1999)+2003=2003