答：a²-5a+1=0两边同除以a：a-5+1/a=0a+1/a=5a²+1/a²=(a+1/a)²-2=5²-2=25-2=23所以：a+1/a=5a²+1/a²=23
a²-5a+1=0当a=0时，原方程不成立，所以a不等于0故可在方程两边同时除以a，得a-5+1/a=0即a+1/a=5(1)a^2+1/a^2=[(a+1/a)^2]-2=25-2=23
