证明：∵Δ=b²-4ac
=(-2a)²-4×(2a-1)×1
=4a²-8a+4
=4(a²-2a+1)
=4(a-1)²≥0
∴无论a为何实数,方程（2a-1）x的平方-2ax+1=0 总有实根.