Question

X，Y属于正实数， X乘以根号1减Y的平方+Y乘以根号1-X的平方=1，求X的平方+Y的平方的值

求过程！！

Answer 1

x√(1-y^2)+y√(1-x^2)=1
[x√(1-y^2)+y√(1-x^2)]^2=1^2
x^2(1-y^2)+2xy√[(1-y^2)(1-x^2)]+y^2(1-x^2)=1
x^2+y^2-2(xy)^2+2xy√[1-x^2-y^2+(xy)^2]=1
1-x^2-y^2+(xy)^2-2xy√[1-x^2-y^2+(xy)^2]+(xy)^2=0
√[1-x^2-y^2+(xy)^2]^2-2xy√[1-x^2-y^2+(xy)^2]+(xy)^2=0
{√[1-x^2-y^2+(xy)^2]-xy}^2=0
√[1-x^2-y^2+(xy)^2]-xy=0,√[1-x^2-y^2+(xy)^2]=xy
so,
x^2+y^2-2(xy)^2+2xy√[1-x^2-y^2+(xy)^2]=1
x^2+y^2-2(xy)^2+2xy*xy=1
x^2+y^2=1

Answer 2

√(1-y^2)+y√(1-x^2)=1
[x√(1-y^2)+y√(1-x^2)]^2=1^2
x^2(1-y^2)+2xy√[(1-y^2)(1-x^2)]+y^2(1-x^2)=1
x^2+y^2-2(xy)^2+2xy√[1-x^2-y^2+(xy)^2]=1
1-x^2-y^2+(xy)^2-2xy√[1-x^2-y^2+(xy)^2]+(xy)^2=0
√[1-x^2-y^2+(xy)^2]^2-2xy√[1-x^2-y^2+(xy)^2]+(xy)^2=0
{√[1-x^2-y^2+(xy)^2]-xy}^2=0
√[1-x^2-y^2+(xy)^2]-xy=0,√[1-x^2-y^2+(xy)^2]=xy
so,