摘要
例1已知(x-1)(x-2)=0,求((x-1)/(x-3)-(x-4)/x)÷(x^2+x-6)/(x^2+3x)的值.解析原式=(x(x-1)-(x-4)(x-3))/(x(x-3))·(x^2+3x)/(x^2+x-6)=(6(x-2))/(x(x-3))·(x(x-3))/((x+3)(x-2)) =6/(x-3),∵(x-1)(x-2)=0且x^2+x-6≠0,∴x-1=0,∴x=1.当x=1时,原式=6/(1-3)= -3.反思在解决有关分式的求值问题时,应注意"分式的分母不为零"这一隐含条件.
