已知x-y=6,根号x^2-xy+根号xy-y^2=9,则根号x^2-xy-根号xy-y^2的值是

Question

已知x-y=6,根号x^2-xy+根号xy-y^2=9,则根号x^2-xy-根号xy-y^2的值是

Answer 1

根号x^2-xy+根号xy-y^2 =s;
根号x^2-xy-根号xy-y^2 =t;
根号x^2-xy+根号xy-y^2=9
[根号x^2-xy+根号xy-y^2]^2 =81=x^2-y^2+2根号[(x^2-xy)(xy-y^2)]
x^2-y^2+2根号[xy(x-y)^2] =81
6(x+y)+2根号(36xy) =81
6(x+y)+12根号(xy) =81
2(x+y)+4根号[xy] =27;
(x+y)^2-(x-y)^2=4xy
(x+y)^2-4xy=36
[根号x^2-xy - 根号xy-y^2]^2 = 6(x+y) - 12根号(36xy)
(st)^2=[6(x+y)+12根号(xy)][6(x+y)-12根号(xy)]
=36(x+y)^2-144(xy)=36*[(x+y)^2-4xy]=36*36
st=±36
根号x^2-xy-根号xy-y^2 = t = ±4