Question

若x^2-3x+1=0,则(x^3+x^2+x)/(x^4+x^2+1)的值为

A.1/2;B.1/3;C.1/4;D.1/5

Answer 1

(X^3+X^2+x)/(x^4+x^2+1)=[X(x^2+X+1)]/[(x^2+X+1)(x^2-X+1)]=X/(x^2-X+1)=X/(x^2-3X+1+2X)=X/(0+2X)=X/2X=1/2

Answer 2

(x^3+x^2+x)/(x^4+x^2+1)
=[x(x^2-3x+1)+4X^2]*[x^2*(x^2-3x+1)+3x^3+1]
=(4x^2)*(3x^3+1)
=4x^2*[3(x^2-3x+1)+9X^2-2]
=4(3x-1)[9*(3x-1)-2]
=4(3x-1)*(27x-11)
=4(81X^2-60x+11)
=4[81(x^2-3x+1)+183x-70]
=48(183x-70)