Question

6．1-$\sqrt{2}$的倒數(shù)是-1-$\sqrt{2}$，|1-$\sqrt{2}$|=$\sqrt{2}$-1，2a^-1=$\frac{2}{a}$．

Answer 1

分析 根據(jù)乘積為1的兩個(gè)數(shù)互為倒數(shù)，差的絕對(duì)值是大數(shù)減小數(shù)，負(fù)整數(shù)指數(shù)冪與正整數(shù)指數(shù)冪互為倒數(shù)，可得答案．

解答 解：1-$\sqrt{2}$的倒數(shù)是$\frac{1+\sqrt{2}}{（1-\sqrt{2}）（1+\sqrt{2}）}$=-1-$\sqrt{2}$，|1-$\sqrt{2}$|=$\sqrt{2}$-1，2a^-1=$\frac{2}{a}$，
故答案為：-1-$\sqrt{2}$，$\sqrt{2}$-1，$\frac{2}{a}$．

點(diǎn)評(píng) 本題考查了實(shí)數(shù)的性質(zhì)，差的絕對(duì)值是大數(shù)減小數(shù)，注意求倒數(shù)要分母有理化．