一道初一数学计算提

m/(x-1)+n/(x-2)
=[m(x-2)+n(x-1)]/(x-1)(x-2)
=[(m+n)x-(2m+n)]/(x^2-3x+2)
=(3x-4)/(x^2-3x+2)
总能成立.则对应项的系数相等
m+n=3
-(2m+n)=-4,即2m+n=4
相减
2m+n-m-n=4-3
m=1
n=3-m=2

a^2+1=3a
两边除a
a+1/a=3

a+1/a=3,
两边平方，且中间一项=2*a*1/a=2
所以a^2+2+1/a^2=9
a^2+1/a^2=7
(a^4+1)/a^2=a^4/a^2+1/a^2=a^2+1/a^2=7