Question

求解一题数学题

设 f(x+y,x-y)=2xy/x²+y² ，则 f(1,y/x)= ?
求解详细过程谢谢 那个小2是平方

Answer 1

f(x+y,x-y)=2xy/x²+y²
2xy/(x²+y²)= 4xy/(2x²+2y²)= [(x+y)²-（x-y）²]/[(x+y)²+(x-y)²]

令m=x+y，n=x-y
则f(m,n)=(m²-n²)/(m²+n²)
将m=1 n=y/x 带入得到
f(1,y/x)= （1-y²/x²）/(1+y²/x²)=（x²-y²）/(x²+y²)

Answer 2

2xy=[(x+y)2-(x-y)2]/2.
x2+y2=[(x+y)2+(x-y)2]/2.
所以f(x,y)=(x2-y2)/(x2+y2)
所以f(1,y/x)=[12-(y/x)2]/[12+(y/x)2]=(x2-y2)/(x2+y2),其中x不等于0，且y不等于0

Answer 3

令u = x+y, v = x-y, 则x = (u+v)/2, y = (u-v)/2。因此2xy = 2*(u+v)/2*(u-v)/2 = (u^2 - v^2)/2,
x^2 + y^2 = [(u+v)/2]^2 + [(u-v)/2]^2 = (u^2 + v^2 )/ 2。所以f(u,v) = f(x+y,x-y) = (u^2 - v^2) /(u^2 + v^2 ),
f(1, y/x) = [1^2 - (y/x)^2]/[1^2 + (y/x)^2] = (x^2 - y^2) / (x^2 + y^2)

Answer 4

设：m=x+y, n=x-y
则：x=(m+n)/2, y=(m-n)/2
f(m,n)=f(x+y,x-y)=2xy/(x^2+y^2)=2[(m+n)/2]*[(m-n)/2]/{[(m+n)/2]^2+[(m-n)/2]^2}
=(m^2-n^2)/(m^2+n^2)
所以：f(1,y/x)=(1-(y/x)^2)/(1+(y/x)^2)=(x^2-y^2)/(x^2+y^2)

Answer 5

设x+y=u，x-y=v，求得x=（u+v）/2,y=（u-v）/2
将上述式子带入f（x+y，x-y）=···，求得f（u,v）的表达式，再令u=1，=y/x即可