Question

若1/x+3x=3则x的二次方/x的四次方+x的平方+1=多少?若1/x-1/y=3则（2x+3xy-2y）/x-2xy

Answer 1

原式=X^2/(X^4+X^2+1)

=1/(X^2+1+1/x^2)

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

=1/(3^2-1)

=1/8

(2)

(2x+3xy-2y)/(x-2xy-y)

分子分母同时除以xy,得：

=[(2/y-2/x)+3]/[(1/y-1/x)-2]

=[-2(1/x-1/y)+3)/[-(1/x-1/y)-2]

=(-6+3)/(-3-2)

=3/5