✔ 最佳答案
對所有實數x,y,恒使x^2+y^2-2m(x-2)-2n(y-3)≧0成立,則(m,n)
的整數解有幾組
Sol
x^2+y^2-2m(x-2)-2n(y-3)≧0
x^2+x(-2m)+[y^2+ 4m-2ny+6n]>=0
D1/4=m^2-(y^2+ 4m-2ny+6n)<= 0
m ^2-y^2-4m+2ny-6n<=0
y^2-2ny+(-m^2+ 4m +6n)>=0
D2/4=n^2-(-m^2+ 4m +6n)<=0
n^2+m^2-4m-6n<= 0
m ^2-4m+4+n^2-6n+9<=13
(m-2)^2+(n-3)^2<=13
Set p=m-2,q=n-3
p^2+q^2<=13
(1) p=0 =>q^2<=13 =>7組
(2) p=+/-1 =>q^2<=12 =>2*7=14組
(3) p=+/-2 =>q^2<=9 =>2*7=14組
(3) p=+/-3 =>q^2<=4 =>2*5=10組
7+14+14+10=45組
x^2+y^2-2m(x-2)-2n(y-3)≧0
x^2+y^2-2mx+4m-2ny+6n≧0
(x-m)^2+(y-n)^2+4m+6n-m^2-n^2≧0
(x-m)^2+(y-n)^2-(m-2)^2-(n-3)^2+13≧0
So,13≧(m-2)^2+(n-3)^2
if n=0,m=0,1,2,3,4
if n=1,m=-1,0,1,2,3,4,5
if n=2,m=-1,0,1,2,3,4,5
if n=3,m=-1,0,1,2,3,4,5
if n=4,m=-1,0,1,2,3,4,5
if n=5,m=-1,0,1,2,3,4,5
if n=6,m=0,1,2,3,4
(m,n)的整數解有45組