Question

已知m^2-5m-1=0,则2m^2-5m+1/m=

注意最后一项是1/m，不是1/m^2

Answer 1

m^2-5m-1=0,
因m不为0，等式两边同除以m得
m－(1/m)＝5
[m－(1/m)]^2＝25
m^2+(1/m)^2=27
m^2+(1/m)^2+2=29
[m+(1/m)]^2=29
m+(1/m)=√29,或m+(1/m)=－√29

(2m^2-5m+1)/m

=m+(1/m)-5
＝√29－5或－√29－5
2m^2-5m+1/m=√29－5或－√29－5

Answer 2

m^2-5m=1
2m^2-5m+1/m^2
=m^2+m^2-5m+1/m^2
=m^2+1+1/m^2

m^2-1=5m
两边同时除以m得
m-1/m=5
m^2+1/m^2=(m-1/m)^2+2
=25+2
=27
∴原式=27+1=28

追问

注意最后一项是1/m，不是1/m^2，能算不？