Question

已知x²＋4y²-4x+4y+5=0求

Answer 1

x²＋4y²-4x+4y+5=0
（x-2）²＋（2y＋1）²=0
x=2，y=-1/2
所以
原式=（x²＋y²）（x＋y）（x-y）（x＋2y）y²÷【（x＋2y）（x-y）y（x-y）（x-y）²（x＋y）²】
=（x²＋y²）y÷【（x-y）（x-y）²（x＋y）】
=（4＋4分之1）×（-1/2）÷【5³/2³×3/2】
=-34/375

Answer 2

x²+4y²-4x+4y+5=0
x²-4x+4+4y²+4y+1=0
(x-2)²+(2y+1)²=0
(x-2)²=0,(2y+1)²=0
x=2
y=-1/2

(x^4-y^4)/(x^2+xy-2y^2)*(x+2y)/(xy-y^2)÷[(-x^2-y^2)^2/y]^2
=(x^2-y^2)(x^2+y^2)/[(x+2y)(x-y)]*(x+2y)/[y(x-y)]÷[(x^2+y^2)^2/y]^2
=(x^2-y^2)(x^2+y^2)/[(x+2y)(x-y)]*(x+2y)/[y(x-y)]*y^2/(x^2+y^2)^2
=(x^2-y^2)/(x-y)*1/(x-y)*y/(x^2+y^2)
=(x-y)(x+y)/(x-y)*1/(x-y)*y/(x^2+y^2)
=(x+y)/(x-y)*y/(x^2+y^2)
=y(x+y)/[(x-y)(x^2+y^2)]
=y(x+y)/[(x-y)(x^2+y^2)]
=[-1/2*(2-1/2)]/{[2-(-1/2)][2^2+(-1/2)^2]}
=[-1/2*3/2]/{5/2*(4+1/4)}
=(-3/4)/{5/2*17/4}
=(-3/4)/(85/8)
=-3/4*8/85
=-6/85

Answer 3

有题可知(x-2)²+(2y+1)²=0
∴x=2,y=-1/2

带入可得结果