Question

13．$\sqrt{1+\frac{1}{3}}$=2$\sqrt{\frac{1}{3}}$，$\sqrt{2+\frac{1}{4}}$=3$\sqrt{\frac{1}{4}}$，$\sqrt{3+\frac{1}{5}}$=4$\sqrt{\frac{1}{5}}$，…觀察下列各式：請(qǐng)你找出其中規(guī)律，并將第n（n≥1）個(gè)等式寫出來$\sqrt{n+\frac{1}{n+2}}$=（n+1）$\sqrt{\frac{1}{n+2}}$．

Answer 1

分析 根據(jù)觀察，可發(fā)現(xiàn)規(guī)律，根據(jù)規(guī)律，可得答案．

解答 解：由$\sqrt{1+\frac{1}{3}}$=2$\sqrt{\frac{1}{3}}$，$\sqrt{2+\frac{1}{4}}$=3$\sqrt{\frac{1}{4}}$，$\sqrt{3+\frac{1}{5}}$=4$\sqrt{\frac{1}{5}}$，…得
$\sqrt{n+\frac{1}{n+2}}$=（n+1）$\sqrt{\frac{1}{n+2}}$，
故答案為：$\sqrt{n+\frac{1}{n+2}}$=（n+1）$\sqrt{\frac{1}{n+2}}$．

點(diǎn)評(píng) 本題考查了二次根式的性質(zhì)與化簡(jiǎn)，觀察發(fā)現(xiàn)規(guī)律是解題關(guān)鍵．