2个回答
展开全部
x^2+√2y=√3
y^2+√2x=√3
两式相减
x^2-y^2+√2y-√2x=0
(x-y)(x+y)-√2(x-y)=0
(x-y)(x+y-√2)=0
x不等于y
x+y-√2=0
x+y=√2
x^2+√2y=√3
y^2+√2x=√3
两式相加
x^2+y^2+√2(x+y)=2√3
x^2+y^2+2=2√3
x^2+y^2=2√3-2
(x+y)^2=x^2+2xy+y^2=2
2xy=2-x^2-y^2=2-(2√3-2)=4-2√3
xy=2-√3
y/x+x/y=(x^2+y^2)/xy=(2√3-2)/(2-√3)=4√2+2√6-2√3-4
y^2+√2x=√3
两式相减
x^2-y^2+√2y-√2x=0
(x-y)(x+y)-√2(x-y)=0
(x-y)(x+y-√2)=0
x不等于y
x+y-√2=0
x+y=√2
x^2+√2y=√3
y^2+√2x=√3
两式相加
x^2+y^2+√2(x+y)=2√3
x^2+y^2+2=2√3
x^2+y^2=2√3-2
(x+y)^2=x^2+2xy+y^2=2
2xy=2-x^2-y^2=2-(2√3-2)=4-2√3
xy=2-√3
y/x+x/y=(x^2+y^2)/xy=(2√3-2)/(2-√3)=4√2+2√6-2√3-4
展开全部
两式相减得
(x+y)(x-y)+√2·(x-y)=0;
∵x≠y,
∴x-y≠0;
∴x+y=-√2;
题干两式相加得:
x^2 +y^2 +√2·(x+y)=2√3;
即 x^2 +y^2 +√2·(-√2)=2√3;
x^2 +y^2 =2+2√3;
x+y=-√2
→(x+y)^2 =x^2 +y^2 +2xy =2
→xy =[2-(x^2 +y^2)]/2=[2-(2+2√3)]/2 = -√3;
则:
y/x+x/y = (x^2 +y^2)/(xy)=(2+2√3)/(-√3)= -2√3/3 -2
(x+y)(x-y)+√2·(x-y)=0;
∵x≠y,
∴x-y≠0;
∴x+y=-√2;
题干两式相加得:
x^2 +y^2 +√2·(x+y)=2√3;
即 x^2 +y^2 +√2·(-√2)=2√3;
x^2 +y^2 =2+2√3;
x+y=-√2
→(x+y)^2 =x^2 +y^2 +2xy =2
→xy =[2-(x^2 +y^2)]/2=[2-(2+2√3)]/2 = -√3;
则:
y/x+x/y = (x^2 +y^2)/(xy)=(2+2√3)/(-√3)= -2√3/3 -2
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