题目先化简,再求代数式(1−1a−1)÷a−2a2−2a+1(1-\frac{1}{a-1})÷\frac{a-2}{a^2-2a+1}(1−a−11)÷a2−2a+1a−2的值,其中a=(2+1)(2−2)+1a=(\sqrt{2}+1)(2-\sqrt{2})+1a=(2+1)(2−2)+1.知识点：二次根式的性质与化简章节：数与式 / 二次根式
