将分母有理化求出x、y
x=(√2-1)/(√2+1)= (√2-1)^2/(√2+1)(√2-1)= (√2-1)^2=3-2√2
y=1/x=(√2+1)/(√2-1)= (√2+1)^2/(√2-1)(√2+1)= (√2+1)^2=3+2√2
则xy=1，x+y=6，x^2+y^2=(x+y)^2-2xy=6^2-2x1=34
于是(x+y+xy)/(x^2+y^2)=(6+1)x34=238
是不是少个Y=?
将分母有理化求出x、y
x=(sqrt2-1)/(sqrt2+1)= (sqrt2-1)^2/(sqrt2+1)(sqrt2-1)= (sqrt2-1)^2=3-2sqrt2
y=1/x=(sqrt2+1)/(sqrt2-1)= (sqrt2+1)^2/(sqrt2-1)(sqrt2+1)= (sqrt2+1)^2=3+2sqrt2
则xy=1，x+y=6，x^2+y^2=(x+y)^2-2xy=6^2-2x1=34
于是(x+y+xy)/(x^2+y^2)=(6+1)x34=238
就是撒,以后仔细些
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