解:f(x)=1/(x²-2x+3)=1/(x²-2x+1+2)=1/[(x-1)²+2]平方项恒非负，(x-1)²≥0(x-1)²+2≥2≠0，无论x取何实数，函数表达式始终有意义，函数定义域为R。1/[(x-1)²+2]≤1/(0+2)=1/21>0(x-1)²+2>0f(x)>00
