当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值
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当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值
当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值
当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值
x^2+y^2=5;
x^2-2xy+y^2=1;
2/xy/[(1/x)-(1/y)]=-2;
最后得到值为1