Question

已知(8x-9)^2+2|9y-8|=0,试求(2xy-x)/(3x-2y)+(xy-2)/(2y-3x)+(x-3)/

Answer 1

平方和绝对值都大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.

所以两个式子都等于0

所以8x-9=0,9y-8=0

x=9/8,y=8/9

所以xy=1

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

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

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

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

=0/(3x-2y)

=0