Question #3dd82

1 Answer
Oct 20, 2017

x^2+4y^2-4x+4y+5=0

rArr x^2-4x+4+4y^2+4y+1=0

rArr (x-2)^2+(2y+1)^2=0

rArr (x-2)^2=0 and (2y+1)^2 =0

rArr x = 2 and y = -1/2

Putting these values after solving following equations :-

Given (x^4-y^4)/(2x^2+xy-y^2). (2x-y)/(xy-y^2) -:(x^2+y^2/y)^2

rArr {(x^2)^2 - (y^2)^2}/(2x^2+2xy-xy-y^2). (2x-y)/{y(x-y)} -: (x^2+y)^2

rArr [(x^2+y^2)(x^2-y^2)]/[2x(x+y)-y(x+y)].(2x-y)/{y(x-y)}. 1/(x^2+y)^2

rArr [(x^2+y^2)(x+y)(x-y)]/[(2x-y)(x+y)]. (2x-y)/{y(x-y)}. 1/(x^2+y)^2

rArr (x^2+y^2)/[y(x^2+y)^2]

rArr [2^2+(-1/2)^2]/[(-1/2)[2^2+(-1/2)]^2]

rArr (4+1/4)/[(-1/2){4-1/2}^2]

rArr (17/4)/[(-1/2)(7/2)^2]

rArr -17/4. 2/1. 4/49

rArr - 34/49