(((x+2y)^2)+1)/(x+2y)=2,x和y属于实数,求(x^2)+(y+1)^2的最小(((x+2y)^2)+1)/(x+2y)=2,x和y属于实数,求(x^2)+(y+1)^2的最小值

Question

(((x+2y)^2)+1)/(x+2y)=2,x和y属于实数,求(x^2)+(y+1)^2的最小
(((x+2y)^2)+1)/(x+2y)=2,x和y属于实数,
求(x^2)+(y+1)^2的最小值

Answer 1

答：
[ (x+2y)²+1 ] /(x+2y)=2
(x+2y)²+1=2(x+2y)
(x+2y)²-2(x+2y)+1²=0
(x+2y-1)²=0
所以：x+2y=1
所以：
x²+(y+1)²
=(1-2y)²+(y+1)²
=1-4y+4y²+y²+2y+1
=5y²-2y+2
=5(y-1/5)²+9/5
>=9/5
所以：最小值为9/5