Question

已知m^2-5m-1=0,求2m^2-5m+（1/m）这道题能做吗

Answer 1

已知m^2-5m-1=0,
m^2-2*2.5m+2.5^2-2.5^2-1=0,
(m-2.5)^2=7.25,
m-2.5=√7.25,或 m-2.5=-√7.25,
m1=2.5+√7.25, m2=2.5-√7.25,
则
2m^2-5m+1/m^2=2*(2.5+√7.25)^2-5*(2.5+√7.25)+1/(2.5+√7.25)^2=
=2*(6.25+5√7.25+7.25)-(12.5+5√7.25)+1/(6.25+5√7.25+7.25)=
=12.5+10√7.25+14.5-12.5-5√7.25+1/(13.5+5√7.25)=
=14.5+5√7.25+(13.5-5√7.25)/(182.25-181.25)=
=14.5+5√7.25+13.5-5√7.25=
=28;

或者，当m=m2=2.5-√7.25,
2m^2-5m+1/m^2=2*(2.5-√7.25)^2-5*(2.5-√7.25)+1/(2.5-√7.25)^2=
=2*(6.25-5√7.25+7.25)-(12.5-5√7.25)+1/(6.25-5√7.25+7.25)=
=12.5-10√7.25+14.5-12.5+5√7.25+1/(13.5-5√7.25)=
=14.5-5√7.25+(13.5+5√7.25)/(182.25-181.25)=
=14.5-5√7.25+13.5+5√7.25=
=28;