Question

两道关于整式分式的问题.已知m^2-2m-1=0, n^2-2n-1=0求m/n+n/m的值.已知x^2/(x^-2)=

Answer 1

1,m^2-2m-1=0,n^2-2n-1=0

当m=n时,m/n+n/m的值为2

当m≠n时,m,n是方程x^2-2x-1=0的两根,故m+n=2,mn=-1

∴m/n+n/m=(m^2+n^2)/mn=(m+n)^2-2mn/mn=-6

2,[1/(1-x)-1/(1+x)]除以[x/(x^2-1)+x]

=2x/(1-x)(1+x)÷x^3/x^2-1

=2x/(1-x)(1+x)*(x+1)(x-1)/x^3

=-2/x^2

∵x^2/(x^-2)=1/1-根号3-根号2

∴(x^2-2)/x^2=1-√3-√2

∴1-2/x^2=1-√3-√2

即-2/x^2=-√3-√2

故原式=-2/x^2=-√3-√2