x²+√2y=√3.1
y²+√2x=√3.2
1-2得:
x²+√2y-(y²+√2x)=0
x²-y²+√2(y-x)=0
(x+y)(x-y)-√2(x-y)=0
(x-y)(x+y-√2)=0
∵x≠y,∴x-y≠0
∴x+y-√2=0,x+y=√2
又1+2式得:
x²+√2y+(y²+√2x)=2√3
x²+y²+√2(x+y)=2√3
(x+y)²-2xy+√2(x+y)=2√3
∵x+y=√2
∴(√2)²-2xy+√2*√2=2√3
∴xy=2-√3
x²+√2y=√3.1
y²+√2x=√3.2
1-2得:
x²+√2y-(y²+√2x)=0
x²-y²+√2(y-x)=0
(x+y)(x-y)-√2(x-y)=0
(x-y)(x+y-√2)=0
∵x≠y,∴x-y≠0
∴x+y-√2=0,x+y=√2
又1+2式得:
x²+√2y+(y²+√2x)=2√3
x²+y²+√2(x+y)=2√3
(x+y)²-2xy+√2(x+y)=2√3
∵x+y=√2
∴(√2)²-2xy+√2*√2=2√3
∴xy=2-√3