Question

已知x²+4y²-4x+4y+5=0，求下面式子的值

已知x²+4y²-4x+4y+5=0，求下面式子的值

Answer 1

x²+4y²-4x+4y+5=0
x²-4x+4+4y²+4y+1=0
(x-2)²+（2y+1）²=0，
所以x=2，y=-1/2
所以原蔽陪腔式
=[(x^2+y^2)（x^2-y^2）/xy]*x(x-y)/y(x-y)*[y^2/乱迅(x^2+y^2)^2]
=(x²-y²）宏衫/xy*(x/y)*(y^2/(x^2+y^2)
=(x^2-y^2)/(x^2+y^2)
=(4-1/4)/(4+1/4)
=(15/4)/（17/4)
=15/17

Answer 2

x²+4y²-4x+4y+5=0

(x-2)²+(2y+1)²=0 因：(x-2)² ≥0,(2y+1)²慎铅≥0
所以有：(x-2)²=0 得：x=2
(2y+1)²=0 得：肆举y=-1/宽雹好2

原式=(x²+y²)(x+y)(x-y)/xy *x(x-y)/y(x-y)*y²/(x²+y²)²
=(x+y)(x-y)/(x²+y²)
=(2-1/2)(2+1/2)/(4+1/4)
=15/17